Pt 18 ot 2 Bots LA bet ahad fey 
Hie yee wipe 
2H \ Ged don tis de sees nd 

ee Red brew PREM eu 
oD Rigen ot 
aeabede eth 


Fr p! ay 
Nite Ae Pew yee cade, eaters (hehe? 
Ct or het petit Brit: ee , skate aly 


e ‘ ? eh gtd te Sand 
phgh BA ae ie ' ti orbebaNinashivad fdr seat Ana erat eh ty a x Hbeiela ers 
Ceehegra ie ah TAD AA LE RRs ror eae ppt el Tat ola eC et Eye ist Eee she eg dl 
a rian a igor CRE einer ay eet, : : Cj le 

eb Uanbels ngs Das fin eon 
' , ae ak J j r 4 ‘| ¥ 
| att V LUA et nites eae tire 
eta Titer de eran oat en een 
Bel iaeey eh Vee aM reat nike ite POPP eat 14D Fane M soe he ebay ae 
te Ue 2 ; aL Mat Me gh ehet kak ab ol A} Hoe: B uy aera y $84 Fi ; 
Ys Ae a Piet 784" 5) bes 7 72 5; UA kee SI at ae sd uy { 
PT ities pat uC RAE AIT a palynnns 
bps ie yh + EPPeEE DD ae : AS p> 1s Py 12454 ae3) 4 
4 Wp ei fi ee oe Parents Al Df bh! nd '  LeN ihe tne : 
, Ary Cplinthy 
in Bee rye Zi Rite ? 3 ‘ ( 


=< 
mit 


i 


sf 


pee 
if 
-: 


S55 
ten 
ig hes 


= 
Gare 


es 
= 


Paws ds bay 
ily 
Hy "4! ia iy 

SAP To Ob tea aia ge fee =o fi } 

Pe RONE MU Ce Peer aetay denen yt kee 

KN, Hee hae abe aE ; 

BEE IM EER RRR biel ae rei aths it Ahi 

Py arity ahah AnH Bos hatte Hh a bee 

ies Che Hy ey 1 

LR YRBE OO Sa) PRM ee tu Oe Oe IB 

shay ee x } 

MW dae ST eT) BD 

A A es fd r) 


= 


ern 

as: 
= 
om 


—— 


ase 


ee 


te 
ae: 
ge 


5 hie 
ep 


nas 
intl 
Py 


iH 


§ 


Fine Te Pts: 1 jes 
EB AM RD NST if 


ee " ‘ iN) j 
ath ‘tid 347, 
TE PGK Jat Ie eH + 
SEPM yy Hitt Le) APL, Bed Ay) 
NAR PIE PINE Hi 3 rack 
‘ Ly if are 
mite ya) 
y ‘ } 


Lame Age 
ALY on 
titan geen} 

3 


Make sire th 
| Vevat 
hi we te st he 
fi 91 8) : DF FO) 8 fe de eae Dey 
PED SA A ah 5 a bess pe SE: 

Pic te 4, cat) Mey Radiat i! 
) 1771) Fede EW Ete: 
Dhe Pict -) 


Ate 


Je 
raha 


ee = 


ih 
WT 38 § 
Aue 


ee 
ae 
= 
Se 
~ 
Ss a 
= 
x 


Delage 
Rite 


aa 
Face 
3 C2 
ee 
ae. 


4 
He 
weer tii ph $Fi : ny A ibeetg 
bP TC Vy ¥ | ae 
WRAY ey “43 he 7 
uke) Rp paeea iy bie 
! Loe ca ylp tebe oe 
FPR AL ee FLD Fh pete se SBE A 
8 Wy PWR EI RUM Hat S Jedeaetie é 
" RYU teary hy Cee 
ag i 7.) I tle 3 f eh, , iy ely ‘ 2 : 
DAUM HRV LAN tat ite motel! ys 
TEM FBP OT he Pte a fh ee ; FAY 4s Ged 
Pe SL Hand MSY TERING: it, 
ea PU eS 9 bs op i Abd 
RT IRS Pas MT if Me! 
MOE hata Maud 
ue deat est fy { 
ay te ‘ Me 
HALA. 1 Mag ys 
we! ype 


eae) a 
etter = he 
DoS Se 


a Rane 
Le Ta ehh Ete eee ae MY 
TA LHL RE RoR Lies 
Lie dhaistiveecesponie: 
Behe aee arith 
} 


. 2 
Eis . 
ae 


tH 


Witak ‘ J so 
YARN ne ¥ ie fags j bs 3 ER Ty ; 
DLIMa AE Inge wat 7 , 7 
Egat 38 a ae tia thy 
Fhe Lad Rot HY Ae S A, 
Ate So Uy } Bune ie 
hi fre 4 iS, 
Pape at ete, Lim 
ay sian a} Be ea 
Tas) STS D ph 38) 
DE) 2) 
Psi yy Vin ‘ 
: Dif ah Vers 
UP De I PC MPM N Many Heh 
WT UD raha sal ys hhh) bee; 
WE NA aE aM Al i bt 


i ste! td 
OM Bae iy 
i . ‘yh ae 
tenth! 


sat! ) 
ae A 


Soar 
“a = 


~ 
2, 
cs 

2 


Ten 


= 


Sr a 


< 


fr 


Ree MO UN CHAN 

“ So ee Eee 
hee Ai ey 

{ 5 Ape Pee ats 


wie’ 
Ve Ps + EMR RE reer ng | 
aS 4 UF Minit mete ei ; Aina } ? = ee ae bars be 
F 4a h UH Ba poe M yt'y : : betas * 
na ige Se al MDs ts DME VRE VEY MERRY 4 FY padi 
Arties att NO CTA sh us LTB Te ed ete a 0, eat et i hs 
ae Me TEN EEE parti hs eS he ae ae a ijk Hi 
AYE RMD Dy Pe NS Ae ae De aD a) Af Tet 
i 4 Ware seve’ Hebe yp yaw & eee oe 
ye pars by ARE PIA bas BS Ved ii 
cat ahyveit PROT OT ai ibe 2 Ae Dh PME On ed 7 D a 
+3 ey dy mae es} Te OAS aR A eatin e Ie ¢ 4 i 
BE Creer eee UT bers tera ls Be Py + 5 
vidy 2 9 be A igh? » Ohi id: ph . a ; 
be Ab eda he ede % 4) PER EY Bt 
eds Ae be yb F : 1 ee 
> i] r 4 bees yi +> a Patni 
ew) 1 tee thi | My : va ie ; Ha he tr x 
ay Pte ite WUE BSD) aU IN NO Sea OSD HS Bae yy Ae) CD ETE 
Malte Palnleatadhe vane pe anid alte RAPALA Bay acDcoc@r Be oct Sek AAT beta and tures ite golly Aa) 
bye. bev Maa ht The eve MA OED MI MED RU oes Me ss Bt Beda Nay ats! Ad hace Vauie 72 : 
i 
: 


aes 


BAW? 


APN ety theme, Wins Nee ay VOM MUA PAS arp ag PIT 

4 us Jah yh ‘es ys 5 she V5 Ra theth] a be the oe Os 

) r Ar yin) Viv bk se et f 
tell wh 


SS 


’ Cah he He yo Lhe ter a pe uy) A x; 
‘wae: hy » We) ne TAY re 
ey wager i ey) 4,4 pA Reh | yy Sir talhtae 
} Meath: > 1 Bagh © ee AA.) 
ye 2.) " ee 
4. We pede wary behary 
Ye ere Be ae ¥ yy 
te H , ) 
' has) te ye 
whe mH ver wh mad » 
hy bald ae t 
A 


ie 


The person charging this material is re- 
sponsible for its return on or before the 
Latest Date stamped below. 


| Theft, mutilation, and underlining of books 
are reasons for disciplinary action and may 
result in dismissal from the University. 


University of Illinois Library 


L161— O-1096 


ee 


Cd as ak 
‘ Nes i i 


=, 


Ae WR (c} 


- 4) ; Me , 


oF a 
Ep at 


—_ 


. 


mn 
bs H 


as 


as Laat } N 


. 
al 


Cenk 


AD 


——— 


ELEMENTS 


OF 


CROMETRY. 


pe SEAS SST PSE DE SF STORER DORIS PR EASELS EY 


—— 


‘ 1 
— 
| Noe 


L As 
ji wi ——— “aN “~A 
| iy 


i. 
s 


ELEMENTS 


OF 


GH Oe Mee Ths 


THEORETICAL AND PRACTICAL: 


CONTAINING A FULL EXPLANATION 


OF THE CONSTRUCTION AND USE OF TABLES, 


AND 


A NEW SYSTEM OF SURVEYING. 


BY 


REY. GEORGE CLINTON WHITLOCK, M.A, 


PROFESSOR OF MATHEMATICAL AND EXPERIMENTAL SCIENCE 


IN THE GENESEE WESLEYAN SEMINARY. 


New-Dork : 


PUBLISHED BY PRATT, WOODFORD, & CO. 


1848, 


Entered according to Act of Congress, in the year one thousand 
eight hundred and forty-eight, by 


GEORGE CLINTON WHITLOCK, 


in the Clerk’s Office of the District Court for the Northern District 
of New-York. 


SSS SSE PGEE NSS CS SRR TE TH BSP GR 


a 
Stereotyped by C. Davison & Co., 
33 Gold st., N. Y. 


PREFACE. 


WHILE some persons are complaining of constant innovation 
in text-books, and others finding fault equally with those in use, 
one scarcely knows whether or not to make any apology in putting 
forth a new work. One thing however, as it seems to me, is 
clear: that in view of the importance which is justly attached to 
elementary instruction, there can be little danger of too greata 
supply of manuals from which an enlightened community may 
select. If new books of geography and grammar, of arithmetic and 
algebra, are not only acceptable to the schools, in their onward 
march of improvement, but even indispensable for giving them life 
and vigor, why should objection be made to attempts in adapting 
the elements of geometry to the wants of the young and to the 
existing condition of instruction? Why should a blind veneration 
for antiquity cause the elements of Euclid to continue in one form 
or other in our schools, when the luminous Grecian himself, if now 
living, would, we dowbt not, no longer employ them without a 
material remodeling in conformity with the mathematical methods 
of the day? 

If boys are to learn that which they will practise when men, why 
should tyros be so long restricted to processes which, as mathema- 
ticians, they will seldom use? 

Is it essential to the acquisition of competent skill in numbers 
that our arithmetics should be filled with examples, and to the com- 
prehension of general principles that our algebras should observe, 
in the development of forms, an unbroken continuity of progres- 
sion? why in the elegant science of geometry should there be 
neither example nor process?’ I am aware that these suggestions 
are not applicable, in all respects, to certain books on geometry 
recently published ; but an elementary work of sufficient fullness, 
yet moderate in magnitude, highly practical, and, consequently, 
progressive in theory and example, is still, I believe, a desideratum. 
I have endeavored to prepare such a work ; how well I have suc- 
ceeded will, of course, be determined by others. Its chief feature 


44-0 
1o4| 


6 PREFACE, 


will be found to consist, not simply in the acquisition of geomet- 
rical principles, but in a regular progression of method, whereby 
it is intended to teach how to geometrize. 

In pursuing quite out to the end of our geometrical studies, as 
well as at the beginning, the synthetic and undevelopable methods 
of the ancients, we acquire little or no power of going alone— 
we get some geometry, it is true, but still remain almost desti- 
tute of that education in analysis which is far more important. 
Why, in investigating the doctrines of forms, should we studiously 
keep out of sight the general principles of quantity, as if no such 
principles existed, when even Euclid himself could proceed but a 
little way without stopping to construct his, to ws, clumsy book of 
proportion, the best and only algebra at his command ? 

Why, when so much labor is saved and greater clearness ob- 
tained, should we refuse to employ an equation like (a+b)? 
=a?+b*+2ab? Shall we have resources at hand and refuse to 
use them because Euclid was poor? Is it shorter, more satisfacto- 
ry, or productive of finer results, to shut up the circumference of 
a circle between the perimeters of polygons than to avail ourselves 


of the simple symbol Ba , when employed as the vanishing ratio 


of the increments of two variables? Has geometry given birth to 
algebra, and shall she reap no advantage from her offspring? The 
succinct and methodical Franceeur quoting Lagrange, says, ‘* Tant 
que l’Algébre et la Géométrie ont été séparées, leurs progrés ont 
été lents et leurs usages bornés ; mais lorsque ces deux sciences se 
sont réunies, elles se sont prété des forces mutuelles,et ont marché 
ensemble d’ un pas rapide vers la perfection.”’* Again the lumin- 
ous Lagrange, in the first of his ‘‘ Legons ;” ‘* Les fonctions dérivées 
se présentent naturellement dans la géométrie lorsq’ on considére 
les aires, les tangentes,’’éc. t 

In accordance with these views, and in compliance with the re- 
commendation of Lacroix, avoiding double methods, that we may be 
ever pressing on in that body of geometrical doctrines that are most 
useful, I have paid much attention to the classification, endeavor- 


* So long as algebra and geometry were separated, their progress was slow and 
their application restricted; but when these two sciences became united, they lent 
each other mutual aid, and advanced together with rapid pace towards perfection. 

t Derived functions present themselves naturally in geometry, when we consider 
areas, tangents, &c. 


4 
*, a 
oe 
t 
iss 3b 


PREFACE. 7 


ing, consequently, to arrange the subjects in a natural order, so as 
to fall easily into families and readily develope each other. Thus, 
by the simple method of superposition, instead of a long, mixed, 
and circuitous route, the doctrines of parallels are presently arrived 
at, and, as a consequence, all the elementary theorems relating to 
angles and independent of the length of lines, are embraced in a 
first section of moderate length. 

The comparison of equal figures follows ; then that of propor- 
tional lines, which prepares the way for the investigation of areas— 
the doctrines of the circle are not considered till afterwards. Thus, 
in the first part, the topics are kept distinct, and, it is believed, in 
their natural order, by which means the progress is rendered more 
easy and rapid, and the methods of geometrizing are introduced, 
one after the other, as required by the gradually increasing diffi- 
culties of the growing subject. 

For instance, in the first section little or no artifice is employed, 
and the simplest algebra, amounting to scarcely more than the 
common symbolical notation, is sparingly introduced in the second, 
while in the third the algebraic requisitions are somewhat increased, 
especially in the exercises. The method of incommensurables de- 
veloped as a part of a system in the introductory book, is employed 
for the first time in the third section of the second, or first zeomet- 
rical Book; the correlation of figures and change of algebraic 
signs find application in the more advanced propositions of the cir- 
cle in the third Book; and the ratio of vanishing increments draws 
a tangent to the parabola in the close of the first Part. The ele- 
mentary properties of the ellipse and parabola, being as simple as 
those of the circle and as useful in the study of natural philosophy 
and astronomy, are here introduced. Further, as proportion is 
generally included in our works on geometry, I have thought it ad- 
visable to insert an introductory Book, embracing, in a regular series 
of proportions the first doctrines of algebra, as being convenient for 
reference to those already acquainted with the science, and indis- 
pensable to others, who, by taking up these principles as required, 
may wish to proceed in the same class. ‘The first Part, consisting 
of a hundred and twenty pages, is designed to embrace, in theory 
and practice, such an introductory body of elementary geometry— 
all the more difficult problems relating to perimeters or areas being 
postponed—as is required, not only to enter successfully upon the 
study of the higher investigations that follow, but for furnishing, in 


im PREFACE. 


some measure, with tangible and useful matter, those who want the 
disposition, lack the time, or have not the ability to proceed further. 
The first Book of the second Part consists of an elementary system 
of functions, depending on a single variable, and presented constantly 
under the simple notation. It embraces the binomial and logarith- 
mic theorems, Every teacher must have observed that isolated 
methods, like those employed by Legendre in cases of incommen- 
surability, make only such an impression upon the mind as to 
leave a sort of confusion always hanging about them, while that 
which forms part of a system readily commends itself to the under- 
standing, and, consequently, remains ever after a permanent part 
of our appropriate knowledge. Laplace has well said, ‘+ Préférez, 
dans |’ enseignement, les méthodes générales; attachez-vous a les 
présenter de la maniére la plus simple, et vous verrez en méme 
temps qu’elles sont toujours les plus faciles.”* The method pur- 
sued in this book has been judged not only the most perfectf in it- 
self, but, as will frequently happen when connected subjects, instead 
of being disjoined, are permitted to fall naturally together, at the 
same time the easiest. But aside from the indispensable matter 
which it contains, the chief object of this book is to prepare the 
way for what follows in the arithmetic of signs, the construction 
of trigonometrical tables, and the mensuration of surfaces and solids. 
In virtue of the course just alluded to, I have been enabled, in the 
second Book of the third Part, to make not only a more than usu- 
ally full development of the trigonometrica] forms with their appli- 
cation in the practical resolution of triangles, but to embrace also 
the quadrature of the circle and ellipse. In the next Book I have 
developed a system of surveying which I regard as peculiarly my 
own. It is true that the theorem for the computation of polygonal 
areas, which constitutes its chief feature, may be substantially 
found in Hutton, yet I have given to it so much of a new form and 
a demonstration at once generalt{ and of the greatest simplicity, and 
extended it in so methodical a manner to the laying out and divid- 
ing of lands, that it becomes altogether another thing. Years of 


* In instructing, adopt general methods; endeavor to present them in a manner 
the most simple, and you will see, at the same time, that they are the easiest. 

t Not all the demonstrations in our algebras are perfect—for instance, the demon- 
strations of the binomial theorem in some school books, the most widely dissemina- 
ted, amount to no demonstrations at all. 

t The demonstration in Hutton is very tedious, and can hardly be said to be 
general. 


PREFACE, g 


instruction have proved, as hundreds of individuals would bear tes- 
timony, that the theorem here given will save, at the lowest esti- 
mate, two-thirds of the labor ordinarily incurred by the rectangular 
method. A further advantage is that, dispensing with a large and 
faulty table altogether, it is far more accurate—the computations 
being executed by aid of the common logarithmic numbers, calcu- 
lated with greater care and usually extending to six or seven deci- 
mal places, and the operation being so ordered that, without any 
additional labor beyond what is absolutely essential to an honest 
confidence in the result, all gross errors, if any exist, whether of 
the field or the tables, are detected, and if these have no existence, 
the smaller and unavoidable ones very much reduced. 

It is hoped that this book, often requested by my pupils, will 
prove acceptable to the schools generally. 

Of the third Part, embracing the mensuration of solids, spherical 
trigonometry, and navigation, time will permit us to say little more 
than that, by the method pursued, we have been enabled, within 
moderate limits, to give a fuller development of these subjects than 
is usually found in our elementary books. 

The modification and extension of Napier’s Rules demands, how 
ever, a brief historic notice. I demonstrated and extended these 
rules by showing: 

I. When A = 90°, a, = 90° — a, B, = 90° — B, C, = 90° — C, 

sinb = cosa, cosB, = tanC, tanec, | 
sinc = cosa, cosC,, = tanB, tand, 
sina,=cosb cosc = tanB, tanC.,, 
sinB.= cosC’, cosb = tana, tanc, 
sinC.= cosB, cosc = tana, tand: 

II. When a= 90°, B, = 180° — B, A, = A— 90°, &c., 
sinB,= cosA, cosb, = tanc, tanC, 
sinC,= cosA, cosc, = tand, tanB,, 
sinA,=cosB, cosC, = tanb, tanc,, 
sinb, = cosc, cos Bh, = tan A, tan a 
sinc, = cosd, cosC, = tanA, tanB,: 

Il]. When c=a, or the triangle is isosceles, 

sind, = tanA, tan(+B), 

sinA, = tana, tan(4d), 
sin(+b) = cosa, cos(4P),, 
sin(4B) = cosA, cos(4d). 


10 PREFACE, 


Having shown the above extension to Mr. Dascum Green, then 
a pupil, he returned soon after, saying that he had not only verified 
my forms, but had obtained better ones, and presented the modi- 
fications of Napier’s Rules as I had extended them, substantially 
as they will be found in the text. 

This rule, as now extended and modified, possesses a greater 
simplicity and symmetry, and will enable us, in spherical astrono- 
my, frequently to dispense with complicated figures. 

I have added a small set of tables, extending to seven decimal 
places, calculated to answer the wants of the student while pursuing 
the work, and to make him more ready in using tabular numbers, 
by compelling him to interpolate by second differences. After- 
wards he will find it decidedly to his advantage to possess himself 
of the tables recently published by Professor Stanley. 

In conclusion, the advantages which we have endeavored to se- 
cure are: 

1°. A better connected and more progressive method of geome- 
trizing, calculated to enable the student to go alone. 

2°, A fuller, more varied, and available practice, by the intro- 
duction of more than four hundred exercises, arithmetical, demon- 
strative, and algebraical, so chosen as to be serviceable rather than 
amusing, and so arranged as greatly to aid in the acquisition of the 
theory. 

3°. The bringing together of such a body of geometrical know- 
ledge, theoretical and practical, as every individual, laying any claim 
to a respectable education or entering into active life, demands, 

4°. The furnishing to those who may wish to proceed on in 
mathematical learning, of a stepping-stone to higher and more ex- 
tended works. 

How well we have accomplished our object it is not, of course, 
for us to say. We have endeavored to render the work as mechan- 
ically correct as possible, but, residing at a distance from the place 
of publication, we can hardly expect that it will be entirely free 
from typographical imperfections. 


G. C. W. 
Genesee Wesleyan Seminary, June, 1848, 


CONTENTS. 


PART FIRST. 


ALGEBRAICAL PRINCIPLES—PLANE GEQMETRY DEPENDING ON THE RIGHT LINE, 


1. Definitions, Mathematics, quantity, proposition. ........... esses oa oh 
My daxplanation Gf the Sighs te ya, | 3-g: Ooae soe 63's deh eered sai sihmninie 9 21 
Shs A SEOUE BDO COCQUATIOR, a oe Saleikis abu sigergn du'§'on 4: SnaP 405-0 nin boedie fod 23 
4. Inversions of additions, subtractions, Gc... ...4 ccs sees cesessce sees OD 
Go Gorticients aided and subtracted, wots edo css acs neh sme sencas medias 27 
G, Palynomiials..<. ., x... «+ dtd de Seti «Pte IENs 6 wks espe ewe se denndl 29 
7% Changing the sign of a factor —POWersy... wsie0 os dec aided cdescseonces 30 
8. Square of polynomial, binomial, residual —Product of sum and dif- 
TERROR SOs oe ctnal etn ieee eee ed otis aig dmaiawse se eaen ae ews 6 ccmaee 
‘9. Changing the sign ofa -polynomiabiissinn .) eid ved 8. .ee cae ecewsves 32 
10. Degree of product, multiplication and division of powers............. 33 
il. Multiplying or dividing dividend or divisor... ...... 0% as'osssevccescnce 34 
12. A fraction an Expression Of sCiVISIONs woe thos orcs ce sawe's aes 8 sas» os 35 
13. Multiplying or dividing numerator or denominator... .........002...- 35 - 
14, Multiplying a fraction by its denominator... .... 2... cee cose cece cece 35 
PG, BYACtiGis MUS Oe LOPCINET, ono ca costs sense 54 wham Aap sees tncsiac ts 35 
16, Involiiign and evolution of fractions... 40. ass 2240 0th ose sans 56 35 
i, Divasion Dy @ Traction—Teciprocal J. 2 Fs ences ered rede code a ean a's os 36 
Ls. Reduetion® to given dencminetor iss i's nce. os ogls ss syccinia sme cieme’ od ts 36 
19, Addition and subtraction Of 1ractions.... . ov. 2.50 coe0 ener osee woesie She 36 
20. Kractions cleared of subdenoniimators. «eo. os os nsye scee tees seccwesse 37 
91. Schole sasions. common factors, GC... wns none oss cone ues cves sete 37 
2a, Equations: Defined ——Mentical ree. 5 os ras es aeay sees so ness coue seas 37 
23. Transposition —Simple Equation, how solved,...... casey sas «pe dee 38 
SA LAAT ORSG 6a Ov ee Sages be eee bs as phen a4 eb ece'ey + eee 009 sbmiel tage 39 
Section II. Exponents—Proportion— Variation. 
i, Retiprocalepoweisk. ais ox hoigs ch hwnd. clnweelegeies. cds. 40 
Root. of 2) powerextensionics . cicsieden gat denenvacccdcmemmieewie. ° Al 
3. Fractional exponents......ver sees Meuysiyene cqanlemgeans Mw eeSe Le SE 


AND ON THE CIRCLE, ELLIPSE, HYPERBOLA, AND PARABOLA. 
BOOK FIRST. 


ALGEBRAICAL PRINCIPLES. 


Section I. Use of the Signs—Fractions—Simple Equations. 


12 CONTENTS. 
Pacs 
4, All real exponents subject to the same rules,........ secs ceseceeevese 42 
5. Ratio defined—proportion—homologous and analogous terms—antece- 
Gents, CONSCQUEDES<\..5 ics 6 tks Wels ald Baan einis «ok Site o Wis ale lel 43 
6. Inversions—compositions—product of extremes and means—involution 
and evolution—equal multiples... 0... .ss0ceseccocsvbosssencecn 44 
a anverse OY reciprocal Proportion ...-...50.. ines secisens 25% oo eee be) ae ae 
8. Continued proportion—geometrical progression .... ..e. cece veces cvden SO) 
9. Variation defined—notation.. Sr fe ged... oe shee 
10. Inversion — composition “Aiwwatntter-—riniinladt. comparison —com- 
bined comparison....... ‘ide meet wile a Say eral gs, Jia oiias + f's's Sanaa 
Section III. Analysis of Equations. 
1. A single equation resolvable into several distinct eee Ex- 
BINPILES Saisie cereale Se SAE ae oo a eae 50 
2. Bie taer ee acid for palgiaesira ae product _ PODS: Sakis SP 51 
3. Classes of biquadratics solvable as quadratics... ........ 00. Su agi ee 52 
4. Rule for putting problems into equation... .... .... eee wees tills dais. vee 
5. Discussion of the two values of the unknown quantity........... ... . 5d 
6. Equation between constants and variables... .... 2... cece cece cece sees OT 
i. Goemicients etiitted sia Bilis eet PAW, wea as eee ee a ocick cans ty ae (Pee 
Bs RePeieea eC Tas aula See Camas Te ee Dey tile oO take a bad Se's cies ewe cteee 
BOOK SECOND. 
PLANE GEOMETRY DEPENDING ON THE RIGHT LINE. 
Section [. Comparison of Angles. ° 
1. Definitions. Geometry—solids, surfaces, lines.. .... .... ceee coos woes 63 
2. straight line, nafure, origin Of NOMON. Of .scgas etme deceavsenenennes , OD 
3. Corollaries—scholium on parallels.......... bo Boge 08 once ae ieee 
4, Applications —Straightedge, paraltel nti ane 64 
&. Sum of adjacent’angles constant, method... . ..'.'.. 0s 4 as00 00s aves, OO 
6. Corollaries—right angles, sum of ni sebht ae lines formite one 
and the same straight line, vertical angles, sum of angles round a 
LET OR ie SS SES en, EE ies 5 
ph ppeireaténe Papntanele, Guiverore Cree 67 
8. Parallels, method, reversion and sneeninae:* bie Prelate sae al : 3 68 
9. Corollaries—the converse, conditions determined by the achaltty of 
alternate angles, secant perpendicular, lines parallel to the same, 
angles having parallel sides, angles of parallelogram ............. 69 
19, Application -—diawih? PAYallelssiys ca dug sir > «eka nev sc ce ees ape hp see 
11. Sum of external angles of polygon........ ‘ RPM Se 
12. Corollaries—sum of internal angles of pores on, Where son. renteen. 
quadrilateral, triangle, sum of acute angles of right angled triangle, 
the external angle formed by producing one side of a triangle. 
SOMO UID 5:5 oye d staves d Win wagidd d atnionspcisnar w aiden ORME re MATES A em. TO 
TB; WXCKCISES, 000s owe erage ttdsivanboine nendanacdcivees MAIO’ a J 


}. Polygons, when equal, how proved, . ..~ sis sined ‘is aoiris wiubatnins omdpecieticd 73 
2. Corollaries—equal triangles, parallelogram divided by a diagonal, 
distance of parallels, diagonals of parallelogram, isosceles triangle 
bisected, equilateral triangles. ........ er ret et ok eee 
3. Relation of angles in a triangle of atieleal sites ee ey Serer 73 
4. Corollaries—the converse, a triangle having two equal angles—three. 73 
5. Relation of two sides of a triangle to the third... ........ 0.0. cc00 cece 74 
6. Consequences—perimeters enveloped and enveloping, the shortest dis- . 
tance from one point to another, shortest distance from a point to a 
OR cs oa ns balay ke A an py sca fee ee nS Cae ae 75 
7. Triangles having two sides of the one equal to two sides of the other, 
each to each, but the included angles unequal.... ........ ..02 cece 76 
8. Consequences—triangles having their sides severally equal, a quadri- 
lateral having its oppositesides equal. fos) 0. Fines cies cede wepes oe 76 
AUR TE MOL CISES., oo n-a594cass 04t 1448 58 eee ese ated dawn 77 
Section III. Proportional Lines. 
I, Segments of linesintercepted by parallels... .. s.00 <6ss neces osee saves 78 
2. Consequences—similar triangles, Sc. 14.2 1... cok cece tee wees sence 80 
3. A right angled triangle divided by a perpendicular........ tase cdee ot 81 
4, Consequences—relation of the perpendicular to the segments of the hy- 
pothenuse, &c., square of the hypothenuse, &C........ ..00 sees voce 81 
5. Relation between the oblique sides of a triangle, the line drawn from 
the vertex to the base, and the segments of the base—consequences. 82 
6. Distance of foot of perpendicular to middle of base................... 83 
7. Line bisecting the vertical angle of a triangle..... 1... eee seeeceeeee 84 
SY Exercises. ,.smecen tian $c lemens oan Vues ogee Wediaes slatsvlenswmevies eau (Oo 
Section IV. Comparison of Plane Figures. 
1, Rectangles—consequences, measure, right angled triangle............ 90 
2. Trapezoid—consequences, measures, eas triangle, compar- 
isons, equalities........ acme s = oe 
3. Triangles having an pee? of hese one y ede 8 an angle of ‘the eee 
CONSEQUENCE. «<2, casera sed dase MER UL HK 6 5c» seine nels oe eeits 93 
4. Bre geas oh. sais hc OMe ne oe awe Tae kok a8 N aee eee code seins oa HOw 
BOOK THIRD. 
PLANE GEOMETRY DEPENDING ON THE CIRCLE, ELLIPSE, 
HYPERBOLA, AND PARABOLA. 
Section I. The Circle. 
1, Definitions—consequence.. Bing « ave no ccctgnmencn veal. LOL 
102 


CONTENTS, 


Section II. Equal Polygons—First Relations of Lines and Angles. 


2. Angles at the centre of acnfel iphone oneeqdencks, MeCASUPES. v0 ue « 


CONTENTS. 


3. Inscribed angles—consequences, ..., taNngent....seeececececccsecceee 104 

4. Angle embraced by intersecting secants measured... .........002.0-- 105 

5. Principle—correlation: of figures... 2... es eek cece Sees cece cece cee te 106 

6. Products of the segments of intersecting chords—consequences........ 107 

7. Product of the three sides of a triangle, how related to the diameter of 

fre circumscriping CHClEs. 13 Suse kuee eee eee ets oe eek epee Sear 108 

*G. Equation of thecircle and’ consequences esi. 4 We Sees tees 109 
Be) SOS ELCRSCR Sete e Tes Sale ow Pin nes bs ro eee e Mae ents seteete toe scale Se Renee 112 

Section Il. The Ellipse. 

1. Equation of the*ellipse ‘and ‘consequences "1. So Siva. ges ot. ese ese. L1G 

am Trangent and conseqnemces’*, 71.4". tue i. ve totes vee Sine eRe ete dee bee 118 
Ss. Npreral and Corollary sec ves cteeen so VE Sade bees s cu ene stirs see ee RD 
Se FGMETCISES TSS Sl STAC IT ee ue cacy THUR Fads Howe cede haste aree cst eens 120 

Section III. The Hyperbola. 
t. Equation-of thehyperbola ss. 250 6. dswd ee tote ced searasee WENGE 121 
Section IV. The Parabola. 

By EQUATION ANG CONSEQUENCES 2 v.04 xen dio ons aan tow ausiend os ud nse manent oe 

2. Tangent—method, consequences. .... 262. sees cece voce best wee: shad 123 
3. Scholia—signification of the symbol ba bial FR ARTS SOP ISUP BOC 124 

4. Ellipse, spies and PRR how related.. fa tan mach sassy Loa? 
i ROC Ses, Loins ass i oe Binet neal Nahe thao Ck aicated aan aaa itene 2 LET 


PART SECOND. 
ANALYTICAL FUNCTIONS, PLANE TRIGONOMETRY, AND SURVEYING. 
BOOK FIRST. 
ANALYTICAL FUNCTIONS, 


Secrion I. Primitive and Derivative Functions of the Form xm, 


4. Detihinonemnd SyDibols 6.2. -0:.4 segs fyssgecen ee seRe ack eae ks | ee 131 
2. Derivativeol y = f& = Go 4 aD aaa A onc ccs cece ce wane uns 133 
3. Derivative of the function, y= Ax‘, a being +, —, &e............... 135 
4, Derivative of y = fx = Ax* + Bxr>+- Cz +-..., and the converse..... 137 
5. Derivatives of equivalent functionsa 4... 4.5: ew exes wes se. o000 cows doce 138 
Section If. The Binomial, Logarithmic, Interpolating, and 
Exponential Theorems. 
1, Binomial Theorem, or development of (a+ 2)... cc. cece cence cece 139 
a. aozariinms, rules of Gperalon.....-.%. «sis seme ssoc cote atin Gee e 145 


CONTENTS, 15 
Pace 
3. Development of y = fz, when aY = x, computations. ........ ..00000. 146 
& Interpolation Airis aadeeon tavc ties eeee Sake CeMeb eee Sater LOG 
or Exercises... Gta ites eee ena os chee seas sevens enbeertratitee” LOG 
6. Exponential theorem, or development of y = a7..... 0... eee sees case. 159 
Section III. General Laws relating to the Development of Functions 
depending on a single Variable. 
1. Ratio of the increment of a continuous function to that of its variable 162 
2. Derivative of a polynomial and multiple function..............-.... 163 
3. Use of intermediate and converse functions ........ 2.06 cee. woes Teves 1O4 
4. Derivative ora functional produeh.3 (esses eee bees ens eee LOO 
oO: Powerdfe function, derivatives 2s. FA Cee eee Pewee ene ess 166 
6: Fraction: of fanctions,-defivatives 3. e039. 6.2 voce Oo 8k ceceder ia Lee 
7. Substitutes in finding derivatives............. CN Bie naty Be tbe ve 167 
8, Expansion of a function, Maclaurin’s theorem.... Sass ere ese tOG 
er AXETCISOS sd PGi tl. Cate hal ELM Tate atta MET EG beats th LOO 
BOOK SECOND. 
PLANE TRIGONOMETRY. 
Section I. Trigonometrical Analysis. 

1. Construction and definitions—complementary arcs, sine, &c.......... 170 

2. Sum of the squares of the sine and cosine, the sine an increasing, the 
cosine. decreasing, fUNctiOn « cin nv eaod “204 bainth* vimsmdccschatenen'g? LAL 
3. Tangent, how related to radius, sine, and cosine...................6- 171 
4. Tangent and cotangent, how related................. Si ei ts ile 171 
Ga Secant and tanzent....+ 2s4+.000% «eee sutvhad s 4s she «meh asc ake 172 
6 An Incremental Vanishing ArG..\s i... tiem sthasepenine mas Teer ee 172 
TJ, Sine and cosine, derivatives Of .. sc opi s'ws os ss» afk eam ta satenbene 173 
8. Sine and cosine developed in terms of arc..... 1... 0.2. wee eer one 174 
9. Sine and cosine of the sum and difference of two arcs.. Aapaicaes Lad 

10. Corollaries—sin 2x, cos 2x; 1-+-cos 27, 1 — cos2v; 1 ns sin ape 1 — sin 

2x; (1+ sin 2x)! + (1— sin 2r)#; sin p+ sin g, sin p—sin gq, 

cos p++ cos g, cos gy— cos p;(sin p-+-sing);(sin p — sin q),&c.; sin 
CFE OR ee race thee Vee Beeb arden anes sonnet cantons he's 90% 177 
11. Tan‘(@ +0), cotan (a-b4),”.'. tan Za, Cot 2a, KC... 20. cee case seseces 178 
12. Denominate equations—radius restored..........se- cece cece cece sees 180 
13. Arc developed in terms of tangent, .-. r computed. ..........2--0----. 181 
14. Trigonometrical lines computed .... 0... .5.. cc ee cece ceee cece come ees 183 
15. Logarithntie simes'and tangents... ici ices cae chee cee'e canis vine sen 188 
16. Arc developed in terms of sine, 2. ..:.... cs 0. eens coe cone teen sane cone 190 
17. Trigonometrical equations solved............. ee yr Pe 

Section II. Resolution of Triangles, and Mensuration of Heights and 
Distances 

1, Projection of one line upon another. ........ .c00 cece ce cous sees ceeee 192 


DMD OP dD 


CONTENTS. 


. Fundamental relation between the sides and angles of a triangle....... 193 
. Sum of squares of two sides—corollary..... 1.2. sees cece cecscese sees 194 
. Sum transformed into product—object Of... ....ssee cece cece eeceseee 194 
. Consequences—sin 3A, cos., tanis, SiN A$. 4%) s.0mp oiey ve oils oss bimmenet LOE 
. melation of sides and opposite angles 7......4.49s a5 ees 05s eynagabane LOO 
. Ratiovof sum/‘of.two'sides‘toalifferenceyc. wwii Wess. wes WEE, 195 
. Cases in Plane Trigonometry—applications........... és eee 195 


Section III. Quadrature of the Circle, the Ellipse, and Parabola. 


1.. Circular sector, how measured, ,:.. area Of CITCIE,). « «6:00 be wipe 09 esse den 201 
2. Ratio of circumference to diameter; .-. arcs of similar sectors, .°. areas 

of circles; circles and their like parts, how related............-+.. 202 
3. Incremental vanishing arc of continuOUS CUIVE..... 2... 200. eee woeee 203 
4. Segmental aroa, derivative of; .... ss cama iacaniils dnible ele Wels lenienieietens 204 
5. Area of ellipse—compared, With, CiFcl@.. + uip:sis/} «winise sale  <inp we¥ehenws . 204 
a Ate OF parabolay. 2 3clasns te oe a5 eigisin> ore = "ts 6s kee 5 0 baila canes Sele 205 
7. ‘Proximate-area of CONLINGOUS. CULV Os ssc sidan oa sivicdndueas saenteees apa 206 
As TORETCISON, i. «de! 40s cp os Grea tae e > he aw iamiea wemneNGslbe ss pas aoee ane 


BOOK THIRD. 


SURVEYING, 


Section I. Description and Use of Instruments. 


1, The chain—tlength, division, how used, field notes... .... ..s2 sees see 208 
2. The surveyor’s cross—construction and US€.... 2... cee. cane cece eee 210 
3. The compass—how graduated and lettered, how used............-..- 210 
4 Vernier or nonius......... gecress Ceoe ses cee AEs DNR SSS Eee ee eee 
5. Theodolite—how sptsiinted awed Apes aks peue ase cane <5 pc tevatogy ter. 212 
Sev aridtin Of NRCS tn ata tle eae yt sae fena as ner ha vinemetens fake ree 214 
BAC ELIS eis aati bh waleiwn bee sr GOA Obie Ns aie Fos Op. 34 $4.4 ood aaien oa rate Tae 


Section II. Plotting. 


. Graphical problems—perpendiculars, parallels, &c.......--.-....003 216 
Problems of construction—-% = @ ++), Ge... 2. wie se ew emecngecs srgmes CLO 
~ GtagGation of the circke—-chorgg, . 3.3 od < ceia es arming > do a0 ap ae 


To plot a fel, ‘Diagonal scale Sector copa tire» cemdoen aa neces canen 222 


be ERRORS METAL Oo a a ee he sess clpusimiha bata bigs hin on ghana 224 
nat BTEC PVM a at al es as QUA cid, AW DOs A OM dlin a Aachen ote in ne 


Section III. Computation of Areas. 


1. Last side, and diagonals of polygonal fields..... 6... .... cee wees cone 225 
2. Corollaries. Similar figures and proportional lines; the Pantograph. 227 
Bye LECTCISOS, 5 4s tc k's aie ee AN Sekai AUD. 05:4 bis bccn o>, 9454 4 kip ene 229 
4. Area of polygon in terms of the sides and their inclinations,.......... 231 


CONTENTS, 17 


Pac 
. Corollaries—triangle, parallelogram, similar polygons, equilateral, 


5 
regular—equilateral triangle, square, &C..... 0.2... 0. eee coon eee 232 
© Form of compuanenayniet ees eRe Ae ee a 234 
i. Hercices, pr peasenart series cars eel pee ee tiee Deen eee 236 
©. Dividing-andiwying out lands: cz. cvs. eeaecct ee 12 POAT IN Fe 237 
PART THIRD. 
SOLID GEOMETRY, SPHERICAL GEOMETRY, AND NAVIGATION. 
BOOK FIRST. 
SOLID GEOMETRY. 
. Section I. Planes. 
1. Position of plane, how determined, consequences.........++. sees cee: 245 
2. Line perpendicular to: plane ,COnseQUen Cesta, ta. «Ni o0sd asian cg vs eb ees 246 
@: Parallel planes iniersectetl, Consequences. co. <..ajc.s sae va ee acres gee .. 247 
4, Similar figures described by the revolution of a line passing through a 


fixed point and intersecting parallel planes, cone, pyramid, cylinder, 
Pisin, Helse dee Se ae Pode eee ees sc OM ET eee vecce beleclascecemss 248 


Section II. Surfaces of Solids. 


in Polyhedron, Surface, CONSEQUENCES, <5. .4< 0 cc eo as 35.4 mente dae ead 250 
®.. Surface of revolution, devivalive, -7..4 550 ese ac 8x «> 0d dae yaw mageetodins 251 
ae Spherical zone," comsequenees, cc tess aise ds nape bans.scgetuss nog beean ae cae 
Be PRERCISER I Bia tad ens eee AO Aa oer de haps nee decks SRR A Mad EN a ee 


Section III. Volwmes. 


1. Rectangular anette re how related, consequences. .......... 253 
2c Pyramid, measured: ..... 2.252. sig x's «OOM BEMIS St lore een EBS 
3. Frustrum of cone and paraniia, cinseauenee Fas Cbdatteie e 0g vee sy ais 39 255 
4. Derivative of solid generated by variable plane...............2. c000. 256 
5. Consequences, . phat frustrum, .’. ellipsoid, sphere, ..., Para- 
 poloid.. hi Seis i disk EDR RaW ha we 05 + OT, i eRe ee ee 
6. Similar sole; how palaces, oP RIT SRR Re US Ae 
Fe ESGTCRAE A ede shee wien 6085 6 qe 0 es ETO OU REN oe ede. Sed 259 


BOOK SECOND. 


SPHERICAL GEOMETRY. 


Section I. Spherical Trigonometry. 


.. Sphere defined, cOnsequences, i. cies cere cess ewes tevnceeendes sien OL 


18¢ CONTENTS. 


PAGE 
2. Spherical triangle measured, CONSEQUENCES. .... sees ceee seve cone sence 262 
3. cosa = cosh cosc + sind sinc COSA, CONSEQUENCES. ..+. eee voce cove cers 264 
4, Sides, how related to opposite angles, CONSEQUENCES. .... 202 eee vee 267 
5. Elimination, first, second, siiciaahaitans kh) were. 268 
G> Napier’s analogies 20.26 ¢9<0/<< vo censes+ ives dam sah poe tae eeteeeee See 
7. Napier’s Rules ETM hee igh 's bn eke RN ap Apes «54 lee 273 
8, Cases in spherical trigonometry..... 6... ceee cece cece cece seen ceceees 275 
i EROPCISOG. i aaa cus wee sada s Coes os cea oh pe aes SPte meee 4 cee eevee eee 

Section II. Projections of the Sphere. 

1, Orthographic projection, consequences. .... 2.2. cece eee sees cece cee 280 
RAW TRQOG: «dn. tcis nonce, 5 4's he soe Nba y. 59 ok 8 4.8 eels Steet nie a ea Sw OSs 6 0m ie 282 
St. AJROMORIO PIOJCCLION.. wv s- »cee's¥os ssh case Lnse Gnne ae chvend yUee tute 283 
a, CONIC SECHON.. .5..\ ss «sR RE EGE + SERB REMe 05 54 ee es tips vend aye <> 284 
5. Steréographic projection, consequences, .'. 4 20. v= 0c wsse vepe ewes eves 284 
©. Mow made. ....4'0. 5h deh EO eb MEM ae ce cane ngtiayades 286 
a. Conical ‘projection, hOW Made. as. aye «04 cmiers os ce oeue s650 meee Aeon 288 
SL PACECIBPH, . .. a kno vainh oo ar ere eee eee Ree seas Ones Ac oe She Palys ek ol ee 


BOOK THIRD. 


NAVIGATION. 


Section I. Problem of the Course. 


L. Pitterence of Fatitude 4 ob saces aac be rng cbc Ae oees DOR ep aels ek & 291 
a, Difierence of longitude, consequetices.. oj... <5 5e/swue weSe wees ones a sen 292 
Be aPauer saihinig. 2 3S ee hese Wek Si bee ote eae Weep op knee tate 293 
SS CHOMM NITE N's 2 SEAN. VST PSUS CER i ok oe Sere ety eee chee eee 294 
Jer LERCTCUSCRS 04's St LUGE VLR ee Se Ss oh ee eee ooh eee Ate et Be ee rae 294 
Section II. Problem of the Place. 
De Lavine ‘by meridian AMUUTde. 24 ds vscie ny bagels kde oe é RNP bw oe bu coe 296 
an Litae, consequencesuestp7..utiet Ri .sPeGiaa NCE eel 298 
seLongitude, by. lunar distance .o.aveseseos «4a000saecaweneah olen 302 
Section III. Description and Use of Instruments. 

1, Course Mariner's compass, .¥'.9):93h Gn. LG Waka hea o aerg suas yeas 304 
Me OE DUG HOG... vsk wo decuck wb Strdd'n's wa eenia pula de a) anne aes 304 
3. Zenith distance—sextant, adjustments, use, depression of horizon, re- 


fraction, marallax, semidiamMeter, ...< ness seen: saanaye v4 ROSS 305 


ADDENDA [IME ca es <o00 done os Sdade en ee ere hake ae ena pany keelane 315 
Punine Coeariinine of cuuntes its She Dickie «sO ho arose. 02 waA~ doa 316 


Linparithaiie Paneae. sis as saciveae tan oWese aes hewn cp semmueee 319 
Logarithmic ‘Tangertas aids tech, tbe ROME a <5 cas none ceeree 322 


P Ashe Deabseleh De 


ALGEBRAICAL PRINCIPLES.— PLANE GEOMETRY 
DEPENDING ON THE RIGHT LINE, AND ON 
THE CIRCLE, ELLIPSE, HYPERBOLA, 

AND PARABOLA. 


a: tg Re? 


vel 
, vat 
=T ‘ ‘ 


, 7 . - \ vs s 
(joo aetiain Ps eon metert ili 


és Dait 


HE Ayre eat eae a 
etn ars SREB | yHeT. He, BRRIL eeaA, ale anh 
fas ven edad gomet: ua ORES a bere 


oe 


Bh J yy ane Ll tae 
Ree i" aa Peat a 


ey phe 5 f 
; FY 
eee nance! Pe abe: cahineneasie al 


: P a . re 
* 
v 


BOOK FIRST. 


ALGEBRAICAL PRINCIPLES. 


SECTION FIRST. 


Use of the Signs— Fractions—Simple Equations. 


Definition 1, Mathematics is a science, having for its object the 
investigation of the relations that quantities bear to each other. 

Def. 2. We denominate Quantity that which admits of measura- 
ble increase or diminution. 

Thus, lines, angles, weight and time, are quantities—but color, 
being incapable of a unit of measure, cannot, in the mathematical 
acceptation of the term, be regarded as a quantity. 

Def. 3. A Proposition is anything proposed—if to be done, it is 
called a Problem ; if to be demonstrated or proved, a Theorem ; 
if a direct and necessary consequence of something going before, 
it is a Corollary ; and if evident in itself, or not capable of being 
reduced to any simpler principle, it is known as an Axiom. 

Ecplanation of the Signs. For the investigation of general 
propositions, it will often be desirable, and not unfrequently even 
indispensable, to be in possession of a method, equally general, 
whereby we may denote quantities and the operations to be per- 
formed upon them. The letters of the alphabet are employed for 
this purpose—the first commonly indicating known, and the last 
unknown quantities ; but the student should accustom himself to 
regard any letter either as known or unknown, Instead of using 
different letters, the same differently marked, as a’, read a prime; 
a’ [a second]; @, [a sub-two, or a second], and so on, are fre- 
quently introduced with advantage, 

The symbols of operation for which we shall have more imme- 
diate use, are the signs of 


22 USE OF THE SIGNS. 


Addition, Subtraction, Multiplication, Division, 
-+ (plus, more) — (minus, less) X ore > Or + 


Aggregation, Equality, Inequality, Deduction, Continuation. 


(), 00) 3€0)23 ae SR ETOMIRT), out outs 


Thus, that 4 is to be added to 6, is written, 6-+-4, and read, six 
plus four; that six and four are equal to 12 diminished by two, is 
written, 

6-+-4 = 12 — 2, 

and read, 6 plus 4 is equal to 12 minus 2; that the sum of 6 and 4 
multiplied by 2, is equal to 44 diminished by 4, and the remainder 
divided by two, is written, 

(6 +4) x 2 = (44—4) +2, 
or (6-+- 4) « 2= (44 —4) ;: 2, 
or 2(6 ++ 4) = (44—4) : 2; 
for when the omission of a sign, as is done in 2(6-+-4), would not 
be attended by any ambiguty, it may be dropped. Thus a 5b sig- 
nifies that a is to be multiplied into 5, Propositions are much 
shorter in symbolical than in common language, and are, conse- 
quently, more clearly expressed, as has already been shown, and as 
will appear in a still stronger light by writing an example or two in 
corresponding columns, 


a‘='5, | the a is equal to }, 


c=b; >same ¢ and ¢ is equal to b; 

a=. as (therefore a is equal to c. 
(a+b) (a—-b)| The sum of a and 6 multiplied into the differ- 
“m _\ence of a and 3}, and this product divided by m, 
aa—bb gives a quotient which is equal to the quotient 
Ce a arising from dividing the remainder of the square 
aa of a diminished by the square of b, by m, which 
< m again is less than the quotient of the square 


of a divided by m. 

Scholium. The student should be accustomed to turn the alge- 
braical into common and appropriate language ; thus the sign of 
equality, =, will commonly be read, ‘is equal to,’’ but sometimes, 
*‘ will be equal to,’ and at others, ‘equalling ;” the symbol of 
deduction, .., will generally be read, ‘ therefore,” sometimes, 
‘hence,’ ‘it follows,” and again, ‘from what goes before, we 
infer,” &c.; the symbol of continuation, consisting of three points, 


AXIOMS, 23 


wey Will be enunciated, * &c.,”’ and so on,’’ “ continued according 
to the same Jdw.”’ 


PROPOSITION I. [axtos.] 


The whole is equal to the sum of all its parts. (1) 


This proposition is an axiom, that is, evident of itself; no words 
about it, therefore, can make it any plainer. 

Corollary 1. The whole is greater than any of its parts, (1,) 
or, the whole exceeds any of its parts by those which are except- 
ed—otherwise the whole would differ from the sum of all its parts. 

Cor. 2, Quantities which are equal in all their parts, are (15) 
said to be equal to each other; for the whole is known by its parts 
--or, quantities which are not evidently identical, can be compared 
only by a resolution into like or unlike parts. 

Cor. 3. Quantities which, however resolved, are unequal (1,) 
in any of their parts, are not equal to each other (1;). 

Cor. 4, Quantities which are equal to each other, are (1;) 
equal in all their parts; for, if some of their parts were unequal, 
they would, by (1,), be themselves unequal; .°. 

Cor. 5. Unequal quantities are not equal in all their parts. (1,) 

Cor. 6, Quantities which are equal to the same or equal (1,) 
quantities, are equal to each other; for they are equal in all their 
parts, (1), (Is). 

Cor. 7. Quantities measured by the same or equal quantities, (1,) 
are equal to each other; for equality of measures implies equality 
of parts, whether the measuring quantities be of the same kind with 
those measured or not. Thus, two masses of lead are equal in 
weight when they both contain the same number of pounds, or 
when they both contain the same number of cubic inches. 

Cor. 8. Quantities ate to each other as their measures; .*. (14) 

Cor. 9. Of quantities having ttnequal measures, that is (1, )) 
the greater to which the greater measure belongs. 

Cor. 10. If the same or equal quantities be increased or (1,,) 
diminished by the same or equal quantities, the resulting quantities 
will be equal to each other ; since they will be equal in all their 
parts, (1,), (1). 

Cor. 11. If the same or equal quantities be multiplied or (1,2) 
divided by the same ot equal quantities, the resulting quantities 


24 AXIOMS. 


will be equal ; since multiplication is repeated addition, and divi- 
sion a continued subtraction. 

Cor. 12. If equal quantities be raised to the same pow- (1,3) 
ers, or the same roots be taken of them, the resulting quantities 
will be equal; since a power is formed by continued multiplication, 
and a root is extracted by the converse operation. 

Cor. 13. The same or equal quantities, by the same or (1,,) 
equivalent operations, give the same or equal quantities, 

Cor. 14. Quantities satisfying the same or equivalent (1,5) 
conditions, are equal to each other. 

Cor. 15. Unequal quantities, by the same or equivalent (1,.) 
operations, will continue to be unequal, and in the same sense— 
that is, the greater will be the greater still (1,,). See also(1,,), 
(112), (113). Thus, if a be greater than 4,[a> 6], a increased by 
c will be greater than 6 increased by c [a+ c,> 6-4 c], a diminished 
by c will be greater than 6 diminished by c,[a—c > b-—-c], m times 
a will be greater than m times b,[ma > mb], &c. 

Cor. 16. If inequalities, taken in the same sense, be (1,7) 
added, the result will be an inequality also in the same sense (1,4). 
Thus, if @ be > b and c >d, thena+c>b+d, 

Cor. 17. When inequalities, whose differencesare the same, (1, s) 
are added in a contrary sense, the result will be an equality (1,). 
Thus, if abe as much > db as cis < d, thena+e will =d+d, 

Cor. 18, When inequalities are added in a contrary sense, (1,4) 
the sense of the resulting inequality will be that of the greater (1, ,). 
Thus, ifa > 6 andc < d, then will a +c > 6-+d, provided the differ- 
ence between a and bd be greater than the difference between c and 
dfa—b>d—c,..a>d—c+bha+cec>b+d}]. 

Cor. 19, If inequalities, taken in the same sense, be sub- (leo) 
tracted, the one from the other, the resulting inequality will be in 
the same or a contrary sense, according as the minuend is the 
greater or less inequality. 


PROPOSITION II. [corotrary From 13.] 


Magnitudes which may be made to coincide throughout, (2) 
are equal to each other. 


The magnitudes are equal in all their parts. 
Cor. 1, When one magnitude embraces another without (22) 
being filled by it, the first is greater than the second (1.). 


INVERSIONS, 25 


Cor, 2. Magnitudes measured by the same or equal mag- (23) 
nitudes, are equal to each other (1,). 

Cor. 3. Magnitudes are to each other as their measures (1,). (2,4) 

Cor. 4. Of magnitudes having unequal measures, that (2,) 
possessing the greater measure is the greater. 

Scholium I, It is sometimes convenient to make a distinction 
between equal and equivalent, but the terms will generally be used 
as synonymous, 

Scholium IL. It is obvious that all propositions requiring demon-_ 
stration, must be founded, either directly or indirectly, upon those 
which do not, or on axioms; and hence our first proposition be- 
comes the source of a vast amount of knowledge. 

Def. 4. A coéflicient is a figure employed to show how many times 
a letter is taken; thus, in 3a, 3 is the coéfficient of a, and 3a=a 
+a--+a, A letter may be regarded as a coéfficient, as n in 

na=a+a+a-...[n times], — 

Def. 5. Operations are said to be relatively free when the result 
is the same in whatever order they are executed, the one after the 
other. 

Thus, O, O,, the two parts of a compound operation, are relative- 
ly free when 

O, (O.) =O, (O1). 


PROPOSITION III. 


Additions, subtractions, additions and subtractions, are (3) 
relatively free operations—that is, the terms of a polynomial 
may be inverted at pleasure. 


1°, Additions. That 4+ 3 is = to 3+4 will be evident from 
counting the units into which the two sums are resolvable ; 
thus 4+3=(14+14+1+4+1)+(14+1-4)), 
and 34+4=(1+14+1)+(14+141+4+1); 
Pe Ee 4+3=344. 
So a+b=(1+1+4+1+... [a units]) 4+ (1+1+1+... [6 units]) 
=1+1+4+14..[a@+)] | 
= (1-414... [6]) + (141 PI+s. [a]) = b-ba, 
which was to be proved. 
2°, Subtractions. The remainder arising from diminishing @ units 
first by 6 units and then by c units will be found the same as that 


26 | INVERSIONS. 


obtained by diminishing a first by c and then by 5, ora—b—c=ea 
~c-—9b; for, let a containr+6-++-c units, or 

a=r-+b-+-c, which (1°) =r-+-c+b; 
then (1;,;) a—b=r-+c, subtracting 5 from both sides, 


and a—b—c=r, subtracting c from the last equation— 
or a—c=r-+-b, subtracting c from the first, 
and a—c—b=r, subtracting 0 from the last ; 


.(1,) a—b—c=r=a—c—bd, Q. E. D* 

3°. Additions and subtractions, a+b—c=a—c-+5b; for we 
obviously have the same number of units whether we diminish the 
a+ units by taking the c units from 5 or from a, Q. E. D. 


PROPOSITION IY. 
Multiplications, divisions, multiplications and divisions, (4) 
are relatively free operations. 


1°. Multiplications. We may know that 3 « 4=4- 3 by resolv- 
ing the numbers into their component units, and setting these down 
in an orderly way to count; 4 units. 


wo 
thus, by counting, we find 4 units repeated 3 times, : } 1111 


the same as 3 units repeated 4 times, or 3 « 4=4 3. 
a@ units. 
oe Nati 
has ia: Dae 
So, aunits (=1+1+1-+... (a]) repeated btimes, &|11 1... 
is the same as 5 units (=1-+1+1+.., [b]), repeat- GJ111 
ed a times— os 
or ab = ba; eho ee 


hence (1,2) a@ times 6 units, repeated c times will be equal to } times 
a units repeated c times, but 6 times a@ units, repeated c times, by 
what has just been demonstrated, is = to c units repeated b times a 
times—and in order to repeat c units } times a times, we may mul- 
tiply first by 6 and then by a; for, multiplying by 3 instead of ba, 
is multiplying by a number a times too small, and consequently, 
the product, being @ times too small, will be corrected by multi- 
plying again by a; all which may be set down in symbols thus ;— 
* “Quod erat demonstrandum,” which was to be proved. 


COEFFICIENTS. 27 


ab = ba, 
Ll x3) : (ab)c = (ba)c = c(ba) = cba, 
or abc = bac = cba, 


where it will be observed that a is made to occupy every place in 
the product, and in like manner the same may be shown of } and c 
—and the reasoning may be extended by introducing additional fac- 
tors at pleasure. Q. E. D, for 1°. 

Cor, 1. The factors of a product may be grouped in multi- (4) 
plication at pleasure. 

Cor. 2. Any factor may be regarded as the coéfficient to (4;) 
the remaining factors of the product (4,). 

2°, Division. Assuming any quantity, Q, to be divisible by 
others, as 6 and c, is obviously the same as assuming Q to be 
resolvable into factors, two of which are 6 and c; hence, Q being 
divisible by } and c, denoting the third factor by a, we have 

Q=abc; 

*.(1 py) Q : c=ab, dividing both sides by c, 
and observing that the product abc is divided by c by omitting the 
multiplier c; .*. dividing the last by 6, there results 


(Q w¢).: b= a. 
But (1°) Q = abc = ach; 
i Qa:Ab = ze: 
and (Q.2.b) sea; 
*. (Lz) (Qodeb)uit ci (@ .2.:¢).4,0., .Q- Ey D,. for, 2°. 
3°. Multiplication and Division.—Let Q be divisible by 6, or 
Q=ab; 
rat Q.: b=a, 
and, multiplying by c, (Q : b)+c=ac; 
again Q-c=ab+c=aceb, 
Ps (Q' 2: ¢)uinbeeges 
hence (Q :.b)ec=(Qec): b,. QE. D. for3?, 


and the proposition is proved. 


PROPOSITION V. 


Additions and Subtractions in regard to multiplications (5) 
and divisions, are relatively free. 


1°. Quantities otherwise alike are added and subtracted by (5,) 
adding and subtracting their coéfficients. 


28 COEFFICIENTS. 


3 + 2 
("I~ ~~ 
cl Pe hy i | 
Thus, 3 times 4 and 2 times 4 are obvi- 1 sf Tee ok 
ously 3-++ 2 or 5 times 4, sn tO IG ati | 
3-412.4=(842)-4=5-4, Keane Denies bene ode 
a + b 


ee ee oe ae 

So a times c plus 6 times 1+14+14+..4+14+1+4+1-+.. 
c is equal to a+b times c, 1+1+4+1+..+1+1+4+1-+... 
or ac+bc=(a+ djc; 1t+1+14+..+1+141-4.. 

*, (a+ b)c =ac-+ be, in- f FR Se aes 
verting the members of the 
equality, which obviously ° 
may be done, since it is only asserting hie same proposition in an 
inverted order ; 

* (111) (a +5)c — bc = ac, subtracting bc from both sides ; but, 
as a and b may be any quantities whatever, a+) may be any 
whatever, and we may substitute m for a+ 6, n for 5, and, conse- 
quently, m—n for (a+b) —b=a; 
whence me — ne =(m— nec, (5s) 
or (m — n)c = mc — ne, 

Scholium. The last form, resulting from the hypothesis (5,) 
that m is greater than n [(a-++ 5)>b], must also be adopted when 
m is less than 2; otherwise general symbols for the representation 
of quantities would have to be abandoned altogether, as it will fre- 
quently be impossible, as well as generally inconvenient, to distin- 
guish between m and n, whether m be greater or less than n. And 
in order to thus extend the application of the form, or to make it 
general, we have only to interpret the expressions mc — nc, 
(m—n)c, both when m is greater and less than n : 1°, when 
m >n, we have mc > nc and, consequently, mc — nc, m — n, both 
plus; but 2°, when m < n, we find mc < nc, and mc — nc, m—n, 
both become minus. 

It is easy now to extend the operation to any number of terms, 
whether plus or minus. 

Thus, azr+br—cer+dzr+...=(a+b)e—cr+dr... 
=(a+b—c)r+dr... 
=(a+b—c+d)r-+... 
=(a+ 6—c+d++...)a. 

So the first part of the proposition is proved. 


POLYNOMIALS, 29 


2°. Dividing both sides (1,.) of the last equation by xz, we have 

(az+ bx—cx4+dz+...):xr=a+b-—c+d-... 
= (ar) 2>@-+ (bz): 2—(cr) :.2-+(dzr): 7+... , 
and the proposition is proved. 

Cor. 1. A Polynomial, or algebraical expression consisting (55) 
of many terms, is multiplied or divided by multiplying or dividing 
its terms, 

Cor. 2. One polynomial is multiplied into another by mul- (5,) 
tiplying all the terms of the one into all the terms of the other. 
Thus, (a+b+c-+...) (dg +b,+¢,4+...)=(at+b+ce+...) dg 
+(atd+tet+...)b,.+(a+b+c+...) cg +... =aa, + ba, +cag 
+...t ab,-+ bb, -+cb,.+...+ ac, + beg +cc, +... (5,). 

Cor. 3. The number of terms in a polynomial product, is (5;,) 
equal to the product of the numbers denoting the terms in the con- 
stituent polynomial factors, 

Thus, af’ P..’ P;, P:; :.. (nej, denote polynomials of 4,°b,’c,”..: 
terms, the ae ais being m in number, we have 

P+ P, =P, 4 polynomial of a times b terms ; 

Ay Dg) PM egy! ado ed ach EI ghd 
and generally P,+ P,+ P,«...[m]=Pabc-.., [m]. (53) 

If wemake the number of terms the same, a for instance,in all the 
polynomial factors, or puta=b=c=..., there results, 

Pee, ele ee my Pad... Tiny, 
or CPA rar, ts Os (5,) 

Cor. 4. The number of terms in the mth power of a polynomial 
of a terms, is equal to the mth power of a. Thus, the number of 
terms in the expansion of the sixth power of the binomial r+ y, 
[ (c+ y)*®] will be found to be 2° = 64. 

Def. 6. Operations like those preceding, which may be per- 
formed upon the whole of a polynomial at once, or upon its parts 
separately, are denominated linear. 

Cor. 5. The compounds of the above linear operations, (5,0) 
are themselves linear. 

Thus, if we multiply any polynomial, z+y+2+..., by any 
quantity a, and then divide by 6, we find 


fa(ct+ytet+...)]:b=(ar): b+ (ay): b+(az):b+... 


30 LAW OF THE SIGNS, 


PROPOSITION VI. 


A product made up by the multiplication of additive (6) 
quantities, is itself additive ; and is changed in sign by chang- 
ing the sign of any one of its factors. 

This proposition will become evident by comparing (5;) and its 
extension in (5,) with the first part of (4); for, observing that m 
— m may represent any quantity either plus or minus, by making m 
greater or less than n by that quantity, and that m—n, mc — nc 
are both plus when m is greater than n, [m >n],and both minus 
when m < n,(m—n) c= mec — ne, becomes + (m—n)e-+e 
= + (mc —nc), or +++ =-+ when m> n, and —(m—n)++c 


= — (mc —nc), or — + +=-—, when m<n; but the order of the 
factors may be changed at pleasure, c (m —n)=(m—n)c, .*. + 
°—=-—e-+=-—-: from all which it follows that changing the sign 


of a factor changes the sign of the product into which it enters. 


Thus +2e+ye+2Zz¢..=+ (ry)*+2¢...=-+ (ryz) «.. 


HF LYZ © wey 

and —re+ye4+27e..=— (ry)e +270... =— (ryz) » 
= — LYZ 4 wey 
—Le—YetZe..H=fLYZ ey ' 

and —Zre—¥yre—Ze,,=—TyzZ¢.,.., &., Ke. 


Cor. 1. An even number of minus factors enc? a plus (6,) 
product—an odd number, a minus product, 
Thus, —*—e«—e+—-e.,, [2n]* =+,—+ —e —e,,, [2n+1]=— 

Cor. 2, An even power of a plus or minus quantity is (63) 
plus (6.). Thus (+a)? =+ a’, (—a)? =—a+-—a=-+a’, 

(=: GAM Ae + a2". 

Cor. 3. An even root of a plus quantity is either plus or (6,) 
minus [=+], (6,). Thus %/+ a7"=+a4 

Cor. 4. An even root of a minus quantity is imaginary, (65) 
that is, impossible ; for (6,) it can be neither plus nor minus. 
Thus, in </ —a=+7, ris imaginary ; for, raising to the (2n)th pow- 
er, we have — a= (\/= a)® = (+r)*", =+ r", — =-+ impossible. 


Cor. 5. An odd power of a minus quantity is minus _—_(6,) 
(62). Thus, (— zry>=—2 e—TZe —r=-—2', (—x)?*} 
Be hosts ren ' . 


* Whatever n may be, 2n is an even number. 


SQUARE OF POLYNOMIALS, — 31 


Cor. 6. An odd root of a minus quantity is minus (6,). (6,) 
Thus, "°° —a=—r; for —a=(—r)**'=—r*t',—=-, 

Cor. 7. In products and quotients, like signs[+,+, or—,—] (65) 
produce plus [+] and unlike signs [-+, —, or —, +,], minus [—]. 

For products this has already been shown (6,), and, to establish 
the same for quotients, let g be the quotient arising from dividing 
any quantity D by any other d. 


or DD: a= 0, 
whence (1,2) (D: d) + d=qd, multiplying both sides by d, 
or (4) (D.d):d=D=qd; 
= / + D=+q+-+d, whence +:+=4, 
—D=+9--—d, we tee, 
+ D=—q--—d, we fi-=-, 
—D=—q-+4, eee —:+=-—. Q. E. D. 
Cor. 8. In products and quotients the signs +, —, are (64) 
relatively free (6,). Thus, -+e«—=—+«+4,+:—=—:4. 


PROPOSITION VII. 


The square of a polynomial is equal to the sum of the (7) 
squares of the terms and their double products taken two and two. 
For in (5,) making a, =a, b, = b, cz =¢,... , and arranging, we 
have (a+b+c-+...)? =a?+b?+c¢?+...+ 2ab+ 2ac+... 
+2bc-+... (7) 
Cor. 1. The square of a binomial is equal to the sum of (8) 
the squares of its terms increased by their double product. For 
in (7) making all the letters nothing except a and 6, there results 


(a+b)? = a?-+b?4- 2ab = a?+2ad+b?=a?+(2a+b)b. (8) 
Cor. 2. The square of a residual is equal to the sum of the (9) 
squares of the terms diminished by their double product. 
For changing the sign of 6 in (8) we have 
(a — b)? = a? + (— 6)? + 2a (— d), 
or (63), (6), (a — b)* =a? + b? — 2ab. (9) 


PROPOSITION VIII. 


The product of the sum and difference of two quantities (10) 
is equal to the difference of their squares, and vice versd, the 
difference of the squares of two quantities is equal to the product 
of their sum and difference. 


32 CHANGING THE SIGN OF POLYNOMIALS. 


For in (5,) making all the letters nothing except a, b, do, bg, and 

changing a, into a, b, into — b, we get 
(a+b) (a—b)=a? —B’, 

e i BN (ast bla B). (10) 

Q. What relation does (5,) bear to (7)? (7) to (8)? (8) to (9)? 
(5,) to (10)? Which is the more general (5,) or (7) ? 

Scholium. Forms (7), (8), (9) and (10), are important for their 
applications, and remarkable for illustrating the facility with which 
general truths are discovered by the employment of algebraical 
language. 


PROPOSITION IX. 


A polynomial may be freed from the minus sign, or on (11) 
the contrary subjected to its influence, by changing the signs of 
all its terms. 


Every polynomial, as 3a — 5b + c+ 2d—e, may be, so far as the 
signs are concerned, reduced to the form -+ B— C, representing 
the sum of the plus terms, as 3a-+c+2d, by B, and the sum of 
the minus terms, as — 5b —e, by C. Therefore all cases of 
subtraction will be comprised in this general form, 

A~(+ B-—O), 

where it is required to take the polynomial (+ B—C) from any 
quantity A. Now if from A we subtract B, the sum of the plus 
terms, we have A— B, by which all the terms in B originally + 
become — ; but, in taking the whole of B from A, we have dimin- 
ished A by a part of B, namely C, which ought not to have been 
subtracted, since the true subtrahend, B — C, is only that part of 
B remaining after the diminution of B by C; therefore, A—R 
being too small by C, the true remainder becomes A— B+C, 
on the addition of C—and all the terms embraced in (+ B—C) 
change their signs in (— B+ C), also 

A—(+ B—C)=A-—-B+C, 
or —(+ B—C)=— B+ C, 
or —BiAC=—(4+B-—C,) QE. D. 

Remark, The student should be familiar with the resolution of 
polynomials into factors, not only by the addition and subtraction 
of coéfficients, but by the employment of the theorems under (7), 
(8), (9), (10). 


DEGREE OF PRODUCTS. oo 


EXERCISES, 


1°. What is the sum of fifteen times x and seventeen times zx ? 
152+ 172 = how many times x? 

2°, axr+br=? 3° 2a+b)4+3(a+b)=2 4° a(c+d) 
+ b(c + d) =? 
86°. 1085 —7b=? 6°. 35b—20b=2 7°. ac—be =? 

8°. —ad+bd=—(ad—bd)=?2 9°. mz+nr—rzr=? 

10°. ac—ad- bd — bc = (a — b) (c —d,), how ? 

11°. Resolve a? —b?, a*—b*, a®—d*, a® —B*, at® —H!, 
a” — b*, into their simplest factors, 

12°. Resolve 7? +2241, x?+1-—22, 1—2ry—zr+y)+zrzr 
+ yy, into their prime factors. 


PROPOSITION X. 


The degree of a product is equal to the sum of the num- (12) 
bers indicating severally the degrees of its factors. 


For, as two factors, ab, multiplied by a third, az, give a product 
of three factors, aba,, so a monomial of p factors abc... [ p], multi- 
plied by an additional factor, a,, gives a product of p+1 factors, 
abc... X @g[p-+1], and by a second factor, abc...X dab, [p+2]; 
and generally, adc... < Kdgbocq ...[q] = abc... Xdgbec,...[ p+]; 

abc... [p] X bolo... [Gg] X agbgC3... [7] = ADC. K Aadgly i. 
X agb3¢3-..[p+q+r] 5 
*, in general, abc...[p] X a,b,c, ...[q] X @3b3¢3... [7] X ... = abe 
woe XK AgD,Cy 00 X Ayb3C3...[pt+g+r+...]. (12) 

Cor. 1, Powers of the same letter are multiplied together (13) 
by adding their exponents. 

For from (12), making 0, Cy... Gado,€q) «++ Aaj03,635-.. all = a, there 
results, 

a...[p] X aaa...[¢]K aaa..[7r}] X... = aad... 
(pay r-..), (13) 


or Pea ea se are eT ee, 
Cor. 2. A quantity is itvolved by multiplying its exponent (14) 
by the index of the power to which it is to be raised. 
For, making p, q, r,... all= m and m in number (13), becomes 
Oe Ore os [ ep aati io hid, (14) 
(a”)*=a™. 
3 


34 MULTIPLICATION OF DIVIDEND, 


Cor. 3. Powers of the same letter are divided by diminish- (15) 
ing the exponent of the dividend by that of the divisor. 

For, from (13) we have 

a®ri=—aqdPe a’; 

se art? s g’=@a?=a?r-7 (15) 
or{fp+q=2,9g=y], @ :@=a, 

Cor. 4, When one polynomial is divided by another, the (16) 
highest power of any letter in the dividend divided by the highest 
power of the same letter in the divisor, gives the highest power of 
that letter in the quotient. | 

For (15), divisor a’ quotient a? = dividend a?**. 

Scholium. It will be found convenient in division to arrange 
(3) the polynomials according to the descending powers of a given 
letter. 


EXERCISES. 


1°. Divide a—b by a—b. 2°. Divide a®?— 5b? by a—d, 
3°. Divide a?—b? by a—b. 4°. Divide at —b* by a—b, 


Operation. 
a — b) at — bt (a? + a*b + ab? + 53 
The division is a* — a*b ; Subtract by 
here executed Y aeh changing the sign (11). 
by the powers + a°b — a*b? 
of a (16), + a?b? 
+ ab? — ab? 
+ ab? — bt 
+ ab? — bt 


5°, Divide a> — 6°, a’ — 5S, a’ —B’,..., a*— b”, by a—b. 


PROPOSITION XI. 


1°. Multiplying the dividend while the divisor remains (17) 
the same, or dividing the divisor while the dividend remains the 
same, multiplies the quotient ; 

2°. Dividing the dividend while the divisor remains the same, 
or multiplying the divisor while the dividend remains the same, 
divides the quotient; .*. 


MULTIPLICATION OF FRACTIONS, 35 


3°. The value of the quotient is not altered by multiplying or 
dividing both divisor and dividend by the same quantity. 


Def. 7. A Fraction is an expression of division, and arises from 
an impossibility of performing the operation. Therefore, indica- 
ting the dividend, now called the numerator, by N, the divisor or 
denominator by D, and the quotient or value of the fraction by V, 
we have ‘ 
N: D= V, or N=DV;; hence (17), 

Cor. 1. A fraction is multiplied by multiplying its numera- (18) 
tor, N, or by dividing its denominator, D, [N-R=D-. VR, &c.]. 

Cor. 2. A fraction is divided by dividing its numerator, or (19) 
by multiplying its denominator. 

Cor. 3. A fraction is changed in form, without being (20) 
changed in value, by multiplying or dividing both numerator and 
denominator by the same quantity, 

Cor, 4, If a fraction be multiplied by its denominator, the (21) 
product will be the numerator, 


For (18) the fraction 7 multiplied by d= i = (nd): d= 


d 

Cor. 5, Fractions are multiplied together by taking the (22) 

product of their numerators for a new numerator, and that of their 
denominators for a new denominator, 


A ee : ‘ N QN 
For; times any quantity Q = Q times D (4), = D (18) ; 
N, NN, 
nes At\s Ne eel 
substituting D, for geek D times Dt apa ey (18), 
_ NN, 
Baris = (21), 
multiplying both numerator and denominator by D, (20) ; 
N N,N; _ NN, N;  _NN,N, 29 
DD, Dp07 6 DD, > Do DDD os 


Cor. 6. A fraction may be raised to any power by involving (23) 
its numerator and denominator separately, and consequently, any root 
of a fraction will be found by an evolution operated upon its terms. 

For, making N,, Soak = N, 
and D., D3, ... = D, (22) becomes 
N N NNN... [m] 7 y"=5 SS 
D’D DDD... [my ° 


© noe [0] = DDD...[m] (23) 


36 DIVISION OF FRACTIONS, 


« 


Cor. 7. To divide by a fraction, invert it, and proceed as (24) 
in multiplication, 


Let Q : * be any quantity divided by a fraction ; multiplying 


both dividend and divisor by es (17), we have 


a8 (0.2): (5.2)-(a-2) 32 (a.2) ea 


serge Q. E. D. 


N 
Cor. 8. The reciprocal of a fraction is the fraction inverted. (25) 
For Q : ake Q- =,» becomes 1s a1 oy when Q = 1. 


Cor. 9. Any quantity, whether whole or fractional, will be (26) 
reduced to a given denominator, by being multiplied by this de- 
nominator and set over it. 


D_ QD 
F h — e 1 — e— = a 
or we have Q=Q DD 
-10 5 «100 79° 662 
Th 5 side te eee 6, Pie 3 = ‘662 ; 
us $= “79 = 79 = $= 09 = Too = 00 ~ 0 
7 lo = fi 
and Bath aor ee whence the rule for reducing vulgar 
to decimal fractions. 
Reducing as > ae . to the denominator M, we find 
MN MN, MN, 
D PA D, 


from which it appears that the 


b 


TAL oie Mes 
least common denominator, M, must be divisible by each of the 
denominators, D, D,, D3, ... of the given fractions, or, must be the 
least common multiple of all the denominators, if we would have 
the resulting fractions freed from subdenominators, .-, 

Cor. 10. To add or subtract fractions, find the least com- (27) 
mon multiple of the denominators, to which, as a common denom- 
inator, reduce all the fractions, then add or subtract the numera- 
tors (5,). 

This rule will frequently be superseded by the following : 


SUBDENOMINATORS. 37 


Cor. 11. To free a fraction of subdenominators, multiply (28) 
its terms by the least common multiple of the subdenominators (20), 


el okey “IES ARTE a, (atop 
b P d ‘ 
EUS, - ae gS Aug eM fo wy? where, if we 


would make the subdenominators 0, d, 2, i, k disappear, M must 
be divisible by each. 

Scholium I. The rule for managing the signs has already been 
given in (6,), but it may be well to observe that, of the three signs 
pertaining to a fraction—viz., the sign before the fraction and those 
before its terms—an even number produces plus, an odd number mi- 
nus, and any two may be changed ata time. Thus, 


a ark yee ms ers Ae pea eae Ve oe 
$Ea=th  HatWab4+ost:H=tOe5 
$oat(-:-)=4)=+42=4+4:37=40=-; 
ps ONT ONE tat Gy? fy Uy Heeb a 

—Ta-(b i He-H=5-F=-- =O) =4 
a us 


Scholium II Whenever it may be foreseen that, by performing 
an operation the same factor would be introduced into the numera- 
tor and denominator, it should be suppressed. 


Scholium III, Additions and subtractions will frequently be bet- 
ter performed in part before reducing to a common denominator. 


Def. 8. An Equation is an algebraical expression consisting of 
two members, separated by the sign of equality [=]. 

Equations are of different kinds, 

1°. An identical equation is one in which the members are the 
same, as ad = d. 

2°, An equation of operation has one of its members derived 
from the other—of these we have already had many examples, such 
as mc —nc =(m— n)ec. 

3°. An equation of condition expresses a determinate relation 
that must exist between certain quantities, not distinguished as 
known and unknown, as 4(a+ b) =p—q. 

4°, The word equation more commonly indicates a relation be- 
tween known and unknown quantities, such that the latter may be 


38 SIMPLE EQUATIONS, 


derived from the former, as z+ a= 5, whence, by subtracting a 
from both members, we have x =b— a. 

Equations of this kind are denominated simple, quadratic, cubic, 
&c., or of the first, second, third, ..., degree, according as the un- 
known quantity is of the first, second, third, ..., power. 


PROPOSITION XII. 


Any term may be transposed from one member of an (29) 
equation to the other, by changing its sign. 


For, let A+p=B. 
be any equation, A representing the sum of all the terms in the 
first member, except p, which, it is to be understood, may be either 
+or—, and B those of the second member; then, subtracting p 
from both sides (1,,), there results 
A=B—p. Q.E. D. 


PROPOSITION XIII. 


A Simple Equation will be solved— (30) 

1°. By separating the unknown from the known quantities by 
transposition ; . 

2°. By uniting the coéficients of the unknown quantity in one ; 

3°. By dividing by the coéficient thus formed ; 

4°. By clearing of subdenominators. 


Scholium. An equation may be cleared of fractions, if desired, 
by multiplying it by the least common multiple of all the denomina- 
tors—but the solution will in general be more readily accomplished 
by the rule just given. The exercises marked I. at the end of the 
book, may be here introduced. 


PROPOSITION XIV. 
From n equations of the first degree to eliminaten—1 unknown 
quantities. 


1°. Whenever the conditions of a problem embrace two un- 
known quantities of the first degree, it is evident from what has 


ELIMINATION, 39 


just been said that the equations expressing these conditions may 
be reduced to the form 


az + by=c, 
a2+by=c; 
and .*. Serty=t, 
a c’ 
Tae ay 
whence (+ aaa 5) agen bind 
b b' b b' 


an equation involving but a single unknown quantity, which may 
therefore be solved as above. The method may obviously be ex- 
tended to three or more unknown quantities, embraced in three or 
more independent equations—equations not convertible into each 
other ;—and it will become manifest, by making the elimination, 
that the number of independent equations must equal the number 
of unknown quantities to be determined. 


. From several equations of the first degree to eliminate (31) 
one or more quanitties : 
1°. Make the coéficients of the quantity to be eliminated +1 
in each equation ; 
2°, Take the differences of the equations thus formed. 
Example. Given equations (a) to find z, 
122 —8y-+ 4z=8 (1) 
or-+ y— 6z=—13 (2) (a) 
c+ Ty + 14z = 57 (3) 


re ay he 32 —Q+27=2 (4) 
(2): — 6, —4r—tytz=2 (5) {0 
(3) : 14, Tye + hy + 2= 44, (6) 
BO) Ber Wyant Lg 
(4) — (6), ir — Wy = — 214 (8) 
ey Pe, —WVerty=r (9) (d) 
(8):-—23, —hert+y=% (10) 
(—H-+es)e = ty tt (e) 
ty 3 
He 
Substituting 1 for z in (9), we find 
y=trt+ Ht =2; 


, (4) z=2+2.2—-3+1=3, 


40 EXPONENTS, 


After a little practice it will be found unnecessary to write the equa- 
tions (5) and (d), it being equally easy to pass at once from (a) to 
(c) and from (c) to (e). This method will generally produce the 
greatest amount of cancelation, and, therefore, be the most expedi- 
tious. 

The exercises marked II, at the end of the book may be here 
introduced, 


SECTION SECOND. 


Exponents—Proportion—Variation. 
PROPOSITION I. 


A factor may be carried from one term to the other of a (32) 
fraction by changing the sign of its exponent. 

As in subtracting coéfficients, we must be guided by the same 
rule, whatever may be their relative values, so, by a like extension 
in exponents, from (15) we must always have a” : a" = a”~", whether 


m>or<cn; 


™ ™M « m 
Biren oe Oe on Lard f 
a” a” : a™ q"™ ee (m—n) 9 
Na+-» N 
ce athe Se an el aa Co eee 
D Dear-” 


multiplying both sides by ne and the proposition is proved. 


Cor. 1. The rules for multiplication and division, and, con- (33) 
sequently, for involution and evolution, are the same for minus as 
for plus exponents, both being supposed integral, 


1 
For from Go? = ats 
oa 
1 
and re hata eae 
i 
™ n 1 1 
there results a-" « @-" = —____ = —_ = q~ (rn) —gq-m—n, 


EXPONENTS. 41 


and Or de” ee es Stes = PO rg 


alsg [2 =) a Qh = aOR or (a 2) 8 = a ee a, 
(a-*) "(a *)* a = o Meg = a?" anil 80 GR} 
aS {a7"f = a7"; 
 cdntenbiowiye. 
(a=")" a q? =” 
.. (a’)' =a”, whatever integers p and g may be, + or —, 
and .*. Sart =28/(ay = a? = a™ *9, 
Cor, 2. It will be observed that any quantity affected by the (34) 
exponent zero [0] is equal to unity [1]; since a” : a” gives either 
a° or 1, 


=: — aqar-=m 
(r ™ —@q T x 


= Ain. 


consequently (a@=”)-"= 


PROPOSITION II. 


The root of a power is to be taken by dividing its expo- (35) 
nent by the index of the root. 


For from (14) we have </a”™ = */(a")" =a" =a", 


which becomes Ae a a”'", putting r= mn; and the rule thus es- 
tablished must evidently be extended, for the sake of consistency, 
to the case in which r is not a multiple of n—just as, when we can- 
not execute the division of 3 by 5 we express it as a fraction, 5 2, and 
seek appropriate rules for its management. 


Cor. 1, An exponent may be reduced to a given denomi- (36) 
nator like any other quantity. 


nm Tm 
As a” is equivalent to </a", so a™ must be regarded as a, in 
order to be consistent ; and we are to show that 


n Tm 


a” = Ga 
If not, let z be such as to make 


(a+zr)"=a™, 
then, raising both sides to the mth power, observing that the mth 
power of the mth root is the quantity itself, we have 


™m ™ 
(a+ z)"= (rn ) , which involved again 


gives | [(a+2)")= | (ae) "| 


42 EXPONENTS. 


™ \ rm 
or (14) (a+z)"= (a= ) =a"; 
therefore bv evolution, a+ 2r=a, or r=0; 
ge a" =a; 


and the demonstration is finished, since (33) m, », r may be either 
plus or minus. 

Cor. 2, All real exponents, whether plus or minus, whole (37) 
or fractional, are to be managed by the same rules; addition cor- 
responding to multiplication, subtraction to division, multiplication 
to involution, and division to evolution, 


™ n 


Let a’, a" be any fractional powers, the exponents reduced to the 
same denominator, being either + or —: then will 


m n m+n m n 
aed =a* =a’ ". 
mtn m n 
If not, let (a+z) " =a’ea’, 
a Le T ™ es mm n ™m n 
then(a+z)"1"= é ° ar) = (ar . a) ° (a: ° a) ‘ (esr ° ar) th Ts 
m a m n nm n 
or (4) (a+z)°*"=a" ea" sa’... [rf] Xa" 2 atea’e.. [Tr], 
mY Ff ae fT 
= hr: e (a) —— an e a= que 
‘ a+z=a, orz=0, 
m n m—n ™ n 
and Tiles es l=) Mica Sey hall a 
ae rae ed ha ks ae! Laie 
, Bll FOTO, OFAN Ol anise’ at 


™m ™m m Mm 3 ™m 
= =s 2.— _— 2.— a es 
also [2 =m], are a’=a ",.". (a2) =a "ea’=a4 ’, 


~ and generally (c=) i a 1% (a7) daisies 6A =a. 


t im UA Mm t fa rt 
. ae ge e.| a. (ia ry ° 
os NS a= a) =@'=a , ¢ being either + or — ; 


3 u 
a a"=a" , whatever real quantities wu, v, and r 
may be, 
n 


Note. It follows that a” may be regarded either as the mth root 
it 


of the nth power of a,[(a")”], or as the nth power of the mth root 


of a,{ (ci) |. 


PROPORTION. 43 


PROPORTION. 


Def. 1. Ifa=re, r is called the ratio of a to c, whatever r may 
be, whole or fractional, positive or negative, whether capable of 
being exactly expressed in rational terms or not. Thus, the ratio 
of 6 to 3 is 2, of 3 to Gis 4; of 2 to 3 is +75, 4+ nin...» of 
66/1] ©6 ia s/f — 1. 

Def. 2. If of four quantities, as a, c, a’, c’, a has the same ratio 
to c that a’ has to c’, 
or, if a@=re (1°) 
and | SB Be 5, #27) 
then, a,c; a’, c’, are said to be proportional, or a is said to be to 
casa toc’; and we write 

Aik G. nild, 3 C, 
OF duiatrad + & 
and read ais tocasa toc’, 

.. A proportion is an equality of ratios. (38) 

Def. 3. We call a, c, belonging to the same ratio, homologous* 
terms, while a, a@’,as well as ¢, c, are denominated analogous 
terms; a, ¢, constitute the first, a’, c’ the second couplet. 

Inverting the order of (1°), (2°), 
from a=rc we Pak 0 yh a:6:) 

a =rc \have(a =rc then a :c:: 

Dividing (1°) and (2°) by r, 


1 : Bho" 

from @ =rc yaa CiGire Uy if yo Ghee neue id, 
, or (4°) 

; _Ghave) , 1, hep i 

a=re oper themetidens ¢it a. 


Dividing (1°) by (2°), we havea: a =re: rc’ =c: c'; whence, 
if the ratio of a to a be 7’, that of c toc’ will be the same ; 
inns SN yaa be AY, 

a=re' )have(c=r'c, Wen era. orc, 


( 
voto) eter errata te) 

( 

( 


=} 
a. 
w—vew 
~~ 
OO 
Oo 
~~ 


R 
= 


and (4°) a PIS OS 
there results, 


* 'Ouds = homos = like, Nbyos = logos = ratio=comparison. 


PROPORTION. 


PROPOSITION III. 


If four quantities are proportional, they are proportional 
1°. By inversion of couplets ; 
2°. By inversion of terms ; 
3°. By inversion of both couplets and terms. 
By additions and subtractions we have, 
dt = att 
from (1°), (2°) -=re, | a-a'=r(e=¢); 
; mre, 


=e : : 
i Pes 
if 43032 aC; 
Wskhn ? eahe: Sere rea VS 5 > (BP) 
sata:cte::a—a@:c—c'; 
pee gee n(d +e 
from (5°) pare ( ) 
= pen d ) a= ray 
c= re Assy 
c= fos 
if eee Ca ey 
gdsertattecyste tal is ose (8°) 
 ~ate:a—c:ira te sa—c, 


é&c., dc., &c., .*. 


PROPOSITION IV. 


If four quantities are proportional : 

1°, The sum or difference of the antecedents 
is to the sum or difference of the consequents, 
as either antecedent 
to its consequent ; 


(39) 


(40) 


The sum or difference of the terms of the first couplet 


is to the sum or difference of those of the second, 


as antecedent to antecedent 
or as consequent to consequent ; 


3°. The sums or differences, or both sums and differences of 
the terms, taken in the same order, whether homologous or anal- 


ogous, are proportional, 

The principle in (40), 1°, may be extended thus: if 
a ae ae ee eT 
@t SothuGe 26 es 


Pie catnd) BECoy hwy Cot. Bade AE 


PROPORTION. 45 


or if a spo, 
a= re, 
a’ =rc', &c., &e. 3 
thena+a-+a’+..=r(ete+te’-+...); 
” @+at+a+.i:etoeteoe+t.wi:a:e::aic::a'ie':: 
MOG 7. 


PROPOSITION V. 


If any number of couplets have the same ratio: (41) 
The sum of all the antecedents 

is to the sum of all the consequents, 
as any one antecedent 
to its consequent. 

We should also have 

+ peewee eee PES ec SES isarse rea 3023 &ee 
which may be enounced in words, 


From (1°) a=rc 
and (2°) reversed re) ="; 
we have arc =Trca, 


@iertia se-givesl Farissdeu: 
PROPOSITION VI. 


If four quantities are proportional the product of the (42) 
extremes is equal to the product of the means. 

Cor. To change an equation into a proportion, make any (43) 
two factors into which one member may be resolved the extremes 
and the factors of the other member of the equation the means of 
the proportion ; or, to read an equation as a proportion, begin and 
end the reading in the same member. 


Thus 4.3=2.-6, 
may be read 4;323::62:3) 2:4::3:6, 
or 4:6::2 53, eroe1 4: & 
or 322 5:6 24; e O47 93"! 2, 
or 3:62 25, sass. 4 t en 

From (1°) and (2°) we have 

a” =r"c”, 

and ata erg" : 

se ea zich; 


where m may be any quantity whatever, whole or fractional ; .°, 


46 PROPORTION. 


PROPOSITION VII. 
If quantities are proportional, their like powers, or roots, (44) 
or powers of roots, are proportional. 
Again eel ° me, and ma =r + ma’ 
(1°) and (2°) Ra =Ts now fo yt ste =r 8 Re: 
where m and n may be any whatever, whole or fractional; .°. 


PROPOSITION VIII. 


A proportion is not destroyed by taking equal multiples (45) 
or submultiples of homologous or analogous terms. 


Thus, since 4:1227 8 p24, 


we have 16 : 24: : 32: 48, [eq. mult, of anal. terms], 
and 1 : 3:: 1:3, [eq. submult. homol. terms]. 
From a= TC, 

we have a (lem) =r. +(l+m) ce, 

or atma=r (cme); .. 


PROPOSITION IX. 


If any two quantities be increased or diminished by (46) 
equal multiples or submultiples of those quantities, the sums or 


differences thus formed will be proportional to the quantities 
themselves. 


Def. 4. Four quantities are said to be reciprocally proportional 
when the ratios of the couplets are the reciprocals of each other. 


Thus if “=e G, Cc; a’, c’, are reciprocally 
ping ¢ then : 
and a’=4¢'s proportional. 
But from the same equations, we have 
a=Te, 
é =H 3 


Def. 5. Four quantities are inversely proportional when the 
first is to the second as the fourth to the third. 


Def. 8. Wath 26:¢c320: ds syateynthehid,.d, £; d, aay Ore 
said to be in continued proportion, or to constitute a geometrical 
progression; cis said to be a third proportional to a and 4, and 
6 is called a mean proportional between a and c. 

Since froma: b::6: 0c, &c., we have 


PROPORTION. AT 


rains dy 
rb=c, 
Tc = d, 
Teabky 
multiplying together the Ist and 2d, the Ist, 2d and 3d, &c., there 
results riG@=<¢, 
read, 
ate ee 


l being the mth term ; .°. 


PROPOSITION X. 


In a continued proportion we have the 1st term to the (47) 
3d as the square of the 1st to the square of the 2d, or as the square 
of the second to the square of the 3d, &c. ; and the 1st to the nth 
» as the (n—1)th power of the first to the (n—1)th power of the 
2d, Fc. 
aus Seve es eee Os Oe ee nt! ne 

Cor. The last or nth term is equal to the first multiplied (48) 
by the ratio raised to a power one less than the number of the 
term [nz — 1]. | | 

It will be observed that the series is here supposed to be ascend- 


ing, or thata<b<c...; if it be descending, ora >b>c..., then 
r will be less than unity—and in all cases 


By addition we infer that 
r(a+6+c4+...+h4)=b+c+d+...+1, 


or r(S—l)=S—a; [S=a+b+c+..47] 
_tl—a_ a(r*—1) 
cy ROY hs ee BRE 


PROPOSITION XI. 


se ied Asotin. flere (49) 


terms ratio — one 
_ (first term) [(ratio)" — 1] 
.* ratio — one ' 


48 VARIATION. 


Note. Since from (1°) and (2°) we have 


r 


a a 
—=rand —=r7, and .. —=—, 
y c 


we may write every proportion as a fractional equation—whereby 
the many and all the changes that can be rung on proportions, will 
be reduced to very simple operations on equations. 


VARIATION, 


Def. 7. We shall often have occasion to consider quantities, not 
as proportional simply, but as passing by inapprectable degrees 
through all magnitudes compatible with certain conditions. Such 
quantities are denominated VariasLe—and are represented by 
the last letters of the alphabet, as z, y, z, while the first letters are 
used to indicate quantities regarded as constant, or such as are 
independent of the variables. Thus, that y varies as z, a being 
their constant ratio, is expressed by 

y = as. 

In this equation it is to be understood that z, in passing from 
any one given value to another, is regarded as passing through all 
intermediate values while a remains unchanged ; and that conse- 
quently y changes, taking new values depending upon the value of 
xz, On this account z is called the independent, and y the depend- 
ent variable. Thus, if the rate of interest, 7, be constant, and the 
principal, p, be also constant, we shall have a given sum pr, as the 
interest for one year on the given principal ; then if y be the interest 
for the time z in years, there will result the variation 


y = pre, 


This manner of looking upon quantities, not so much as known 
and unknown, as constant and variable, is as important as it is simple. 


PROPOSITION XII. 
If y = az, 
e 1 1 
7a l= Ae y > 1. €. 


If y vary as x, x varies as y. (50.) 


VARIATION. 49 


PROPOSITION XIII. 
Hy ij ¥ = a2, 

y=r=artxr=(atl)r=(atl)- a os TR, 
If y vary as x, x+y varies as x or ¥: (51) 


PROPOSITION XIV. 
If y= az, 
ine aft a aT &:, 
If y vary as x, y” varies as x”. (52) 
PROPOSITION XV. 
If y= O27, 
vi my=a+ mr; i.e., 
If y vary as x, my varies as mx, (53) 
PROPOSITION XVI. 
y = amz, it follows that 


If y vary as x, y also varies as mx. 


PROPOSITION XVII. 


If ¥y = az, 
and Sab7: 
og y = abr; i. e¢., 
If y vary as z, and z vary as x; then y varies as x. (54) 


PROPOSITION XVIII. 


If Ba gat 
and Du be 2 
Rs: yoo (ack bles ise, 
If y and x vary as z, y+ x varies as 2. (55) 


It will be observed that the above forms embrace, very briefly 
and simply, not only essentially the whole doctrine of proportion, 
but a vastly wider field, by reason of the unlimited number of 
values of which y and z are susceptible. 


SECTION THIRD. 


Analysis of Equations. 


The following is a principle of the first importance in analytical 
investigations : 


PROPOSITION I. 


Certain equations are so constituted that they necessart- (56) 
ly resolve into, and are consequently equivalent to, several inde- 
pendent equations. 

We do not propose, in the present article, to enter into a full de- 
velopement of the boundless resources which this principle affords, 
but simply to illustrate it by examples of such particular cases as 
will be serviceable to us as we proceed. 

Required two numbers such that if they be diminished severally 
by a, 5, and the remainders squared, the sum of these squares shall 
be equal to zero. 

Denoting the numbers by. x, y, we have 

(x7—a)*-+ (y—b)? =0; 

and it follows from (6;) that (x — a)*, (y—b)*, must both be +, 
whether x2a,* y25; but it is obvious that, since neither of the 
terms (z — a)’, (y — 5) can be minus, neither can be greater than 
0; for if either term, (xr—a)? for instance, have an additive value, 
the other (y — b)? must possess the same value and be subtractive, in 
order to satisfy the equation, or that their sum may = 0; whence it 
is necessary that 

(x —a)* =0, and (y—b)?=0; 

x —a=0, and y—b=0, 
or x =a, and y= b, 

As a second example, what numbers are those from which if a and 
b be severally subtracted, the product of the remainders will be = 02 

Representing the numbers by z, y, there results 

(x — a) (y—6) =0; 

.. dividing by y—b, xz—a=0, or r=a, 

and dividing by z—a, y—b=0, or y=), 


* xz a, x greater or Jess than a. 


a 


QUADRATIC EQUATIONS, 51 


Every equation of the second degree, or, embracing no other un- 
known quantities than z* and z, may, by transposing, uniting 
terms, and dividing by the coéfficient of x*, be readily reduced to 
the form 

+1e-2?+2ar=5b; (57) 
understanding that a, b, may be either--or —. In order to find z, 
we observe that the first member of the equation will become a bi- 
nomial square (8) by the addition of a? ; 

of z?+2rat+a?=6b-+ a’, 

-*» (8), (64), r+a=+ ,/ (b+ a’), 
and z=—axt,/(b+a’); .., (58) 


PROPOSITION II. 


To solve a QuapRaTic Equation : 

1°. Reduce it to the form of (57) ; 

2°. Add the square of half the coéfficient of x to both sides ; 

3°. Take the square root of the members, prefixing the double 
sign [+] to the second ; ! 

4°, Transpose. 

It will be observed that lf resolves into two independent equa- 
tions (58), 

x=—a+./(b+a*), and r=—a—./(b+ 0’), 
thus illustrating (56). 
Adding these values of x, we have 
[—a+./ (b+ a*)]-+[—a— J (6+ a*)] =— 2a, 
and their product is (?) 
[—a+ V(b+a*))[—a— (b+0%)]=[— a}? —[ V(b +0%)P 
=a*—(b+a’?)=— 6, 

Cor, In a quadratic of the form (57) the sum of the values (59) 
of x, is equal to the coéfficient of z' taken with the contrary sign, 
and their product to the second member also taken with the con- 
trary sign. 

Queries. What will (57) and (58) become, when a is changed 
into —a? binto—b? When 6 is minus, what must be its value 
compared with a’ in order that the value of x in (58) be impossible ? 
[See (6,.)] Will change of value or sign in a ever render z imagin- 
ary? Why not? 

Scholium I. We naturally inquire if the Cubic Equation can be 


52 ROOTS OF EQUATIONS. 


resolved into three independent equations. Every equation of the 
third degree reduces to the form 
z?+az?+ br-+c=0. (a) 
As we must suppose z to have some value—one at least—let r 
be that value ; then must r satisfy the equation, 


or r?tar’tbrt+c=0 
af, c=—r?—ar’*— bdr, 
hich value of c substituted above, gives 
z?+azr’*+ br 
—r?—ar*—br=0, 
or (x? —r?)+ a(x? — 7°) + d(x —1r) =0; (5) 


iz: nei (z+r)+5=0, 
Hividing by x —r, see examples under (16) ; 
+(atr)r+r’?+ar+b=0, (c) 
Phente [(57), (58)] (a) is resolvable into three equations, giving 
three values for z. 

Comparing equations (a), (5), (c), and observing that (5) is the same 
as (a), we learn that, if r be a root of the cubic equation (a), that the 
equation is divisible by z — r, giving a quadratic (c). 

The student is requested to prove that the equation 

zi+pz?+qz’?+rz+s=0, 
is resolvable into the factors 
xr—a,x—b,x—c, x—d, 
or that (x — a) (x —b) (x —c) (uv —d)=2'+ pri +q2r?*+rzr+s=0, 
d,.D, &,.d,,betng values of 2, or 2 4,9 = 6,7 =r, eo 

Scholium II. Many Biquadratic Equations may be reduced as 
Quadratics ; e. 2. 

1°. When reducible to the form 

(2?-+ Ar+ B) (2?+ A’r+ B’) =0, (60) 
where the conditions are 
zt+(A+ A')r?+(AA'+ B+B')x? + (AB' + A'B)z + BB =0, 
ors=BB,qg=AA 4+ BA B,p=At+A,r=AB4S AB. 

2°, When reducible to the form 

(z?+Ar)’?+B(2?+Azr)'=C, (61) 
where the conditions are 
zi4+ 2Azr*?+ (A?4+ B)zr*+ ABr=C, 
or p=2A, r=AB, g=A?+B, 
3°. When reducible to the form (61,) 


a: a 
reg A em 


PUTTING IN EQUATION, 53 
a 
for Qere—=24; 
x 


a\* a \! 
(2+) +(2+ =) = b+ 2a. 
Example. Given zr‘ + 4x° + 62? 4 4x = 15, to find z. 
2 = 
We have A-ha 2, B= a= = and } FRB, 
(x? + 2x)? + 2x? + 2r)* = 15, 
(x? + 22x)? + 2x? + 2r)'+1= 16, 
z?+2r4+1=+4, 
rtl=+/t4=+4+ 20r=+/—4, 
ase (2B P/E 4 or S TS A/S 4. 
Hd many values has r? What values are imaginary? Verify 
by substituting these values in the first equation. 


20] 


Scholium III, When a problem embraces several unknown 
quantities, it may be solved by representing these quantities in 
different ways; but the elegance and facility of the solution will 
frequently depend upon the notation which we employ. The fol- 
lowing rule, taken from the Cambridge Mathematics, (application 
of Algebra to Geometry,) will be serviceable. 


‘“< If among the quantities which would, when taken each (62) 
for the unknown quantity, serve to determine ail the other quan- 
tities, there are two which would in the same way answer this 
purpose, and it could be foreseen that each would lead to the same 
equation (the signs + and — excepted); then we ought to em- 
ploy neither of these, but take for the unknown quantity one 
which depends equally upon both; that is, their half sum, or their 
half difference, or a mean proportional between them, or, &c.’’ 

Thus, suppose it were required to find two numbers such that their 
sum should be 4 and the sum of their 4th powers 82. Instead of 
taking z and y for the numbers, we may make them both depénd 
equally upon z by putting 

2+ z= greater No., 
2—xz=less No.; 
since the sum would be 4 = greater No. + less No., 
(2+ 2)* =164 3227-4 2427’? + 82? + 2%, 
and (2—2)* = 16 — 327+ Mr’? —8r?+2'; 

oT 2r*+2-24r? 42-16= (2+2)*+ (2—7z)* = 82, 

or zit24r* = 41—16=25; 


54 EQUATION SOLVED, 


xt + 247? + 12? = 25-1 144 = 169, 
z?+12= +413, 
4 4 
z= (£13—12)° =+1, or = + (— 28)’ ; 
1 ak 
greater No. = 2+ 2 =3, 1, 2+ (— 25)", or2— (— 25)’, 


2: 1 
and less No. =2—2=1,3, 2— (— 25)", or 2+ (— 25)”, 
where it must be observed that if 3 be taken for the value of the 
greater number, | must be taken for that of the less, and so on for the 
three remaining corresponding values, 
As a second example, suppose the equations 


4 a 
ety + (2? +a?) +(y?+a*) =8, 
and ae ie 
a 
given to find + any y. We shall make x and y depend equally 


upon the same unknown quantity by putting 


ae: 
Po SS S25 
: ee 
a 
whence eid earn 


which values substituted in the first equation, give 
a a} 2 z a* 2 t 
2h ka Zo a)" + mre = 6, 


+. 0 


=, 
a 


or Belek gink Wick le 1 
z e: ii 
b 
dividing by aand putting 7 Tm to avoid fractions ; 


(4+) (24+1t=n—2-4 
Zz ." Zz’ 


ah 
Girt? Lede oth pe n? —2ng — “24 28 


1 
(Qn 4224 T* _ ns, 
n? 
8 en ecient =— 
"Fe ee 
2 
or z*—2mz=—1, putting m= 2 (3 2)? 


4(nt1) n+l’ 


USE OF THE DOUBLE SIGN, 55 
oe z?*—Qmz+m? =m? —1=(m+1)(m—1), 


z=m+[(m+1) (m—1)]*. 

Scholium IV. From notions derived from the practice of arith- 
metic, the student, on first being introduced to problems capable of 
double solution, is frequently inclined to question the propriety of 
the second value of xz ; and, consequently, to reject or neglect the 
second or minus sign, in taking the square root of a quadratic equa- 
tion, Such neglect must zever be allowed; for the double sign 
(+) is necessary—and, so far from being any thing like an excres- 
cence or deformity in algebraical language, is a symbol of the great- 
est utility, as will appear hereafter when we come to apply the fore- 
going principles to Geometry. The following problem, illustrative 
of the use of the double sign, will suffice for the present. 

Required the point in the line joining the centres of the earth and 
moon where their attractions are equal. 


Let E be the earth, M the moon and C the point , Brn luh 9 
of equal attraction, and let the quantity of matter in © 
the earth be represented by e and that in the moon by eae 
m, the radius of the moon’s orbit, or distance from the earth, by r, the 
distance C M by wu and consequently, C E by r— ux. 


Now it is a principle of physics that the attraction of a sphere is 
proportional to the quantity of matter directly and to the square of 
the distance to its centre inversely, or 


qt. of mat. 


to (dist.)2 ; whence 
Earth’s attraction at C will = eee oe $ 
uy 
Moon’s attraction at C will = = : 
e i Sem = ad 
‘ (r—u)? w? 
3 AGREE Bue 
T—Us U 
of wu /e=tr./m—u(j/m, 
oa Co) 6.24/21 a= 2-7 ./ m, 
+r erm m r 


ee = =, 
WhO wl Pine qctan AS 


m 


56 USE OF THE DOUBLE SIGN, 


But the mass of the moon, inferred from her action in raising the 
tides, is +; of that of the earth ;* 


oo a5, and a/ <= VB= 86; 
m mm 


T 1 1 

~ T8668 °" ~ 966 ~ 766 * 
or the distance of the point of equal attraction from the moon, is 
nearly + ;3; of the moon’s radius, or — 38; of moon’s radius, where 
it remains to interpret the — sign in the second answer. If we sub- 
tract from uw (that is, from C lay off a line in the direction of M) 
a quantity less than w, the point of equal attraction will approach 
M, and the more, the greater the line subtracted, so that, when the 
line subtracted is = CM, w will = 0, and when > CM, as CC, u 
will become —, being = MC —CC’=MC—(CM+ MC’) =—MC’. 
Therefore if u = + 33; + r, gives the point C on the left or this side, 
— gs «7 corresponds to the point C’ on the right or beyond the 
moon. Indeed, had we supposed the point of equal attraction at C’ 
instead of at C, there would have resulted 


u aM» OT 5 1, 


é mM Tr 
a = =e Whence u = : 
Se boo? ita —14,/£ 
' ; m 
i 1 : 1 
or USE aOR) TOG ee St raat 


The two values of w are therefore perfectly applicable, and, in 
fact, the problem would not be completely solved without them. 
The second value has revealed a truth, viz.: that there is a second 
point of equal attraction beyond the moon—which, though not con- 
templated in the statement of the problem, is yet a direct consequence 
of the equation drawn from it ; and which, when once discovered 
by the aid of algebraical symbols, recommends itself to the under- 
standing without them, since the attraction of the earth is greater 
than that of the moon. Suppose, for a simple illustration, the 
masses of E and M were as 9 and 4, and their distances to C’ as 3 


and 2; then their attractions at C’ would be as zs and iS and con- 
sequently equal, : 

As a last example in which a single equation is equivalent to two 
independent equations, we give the following important theorem : 


* Poisson, Traité de Mécanique, p. 258. 


CONSTANTS AND VARIABLES. 57 


PROPOSITION III. 


Whenever a problem gives an equality between constant (63) 
and variable terms, the variables being capable of indefinite dim- 
inution so as to become less than any assignable quantity ; two 
independent equations will be formed, one between the constants, 
the other between the variables. 


Thus, let AtX=B+Y 
represent any equation in which A, B, are the sums of ayy constant, 
those of the variable terms in each member, X and Y, being capa- 
ble of continually decreasing and becoming less than any assigna- 
ble quantity, that is less than any assignable part of Aor B. Then 
will Am Boandak = ¥'; 
for, if possible, let A be greater or less than B; first suppose A > B 
by some quantity D, or A=B+D; 

Ne X= Y-—D, 
or Y=D+ XX, 
whence it appears that Y can never be less than D, even should X 
become actually =0; therefore Y is greater than, or at least as 
great as, an assignable quantity D, which is contrary to the hypoth- 
esis, and this contradiction holds as long as D has any value, or as 
long as Ais > B. Again suppose 

A< B, or A+ D=B; 

af, X—D=Y, 
or A=D-+Y; 
whence X is always greater than, or at least as great as D, and this 
is also contrary to the hypothesis, according to which _X, as well as 
Y, is to be capable of indefinite diminution. Therefore it is absurd 
to suppose A either greater or less than B, or otherwise than equal 
to B; 

* A = B, and consequently X = Y, 
which was to be proved. 

The method of proof here employed is called that of ‘ reductio 
ad absurdum,”’ leading to an absurdity ; where, instead of showing 
directly that the proposition must be true, we show that it can not 
be otherwise than true. | 

As an example, let it be required to find the sum of the repeating 
decimal fraction 


333 &c. ="3-+ 03+ 003-4... [ad infinitum]. 


58 LIMITS, 


Let A = sum of all the terms, 
X = sum of term after the nth; 
then A — X will = sum of n first terms, 
or A— X='3-+ 03+ 003-1... [n terms], 
_ B3((1)"—1] 3 [P= C1} 43 oy 
SACL why Woeull Gales ter Wik odode eed 


But (‘1)" may be made Jess than any assignable quantity by tak- 
ing n sufficiently great, which may be done since the number of 
terms is infinite, Thus, if we take 1, 2, 3, &c., terms, or make 
successively 

Ss ie ee 4, Bh yive 
we have (‘1)"= ‘1, ‘O1, ‘001, «0001, ‘00001, ... ; 


8 and X becoming < any assignable quantity, we have (63) 


Ma: it: 
one ee. 
So for the circulate ‘23, 23, 23, &e., 
we have A — X = ‘23 + +0023 + 000023 +... [n], 
23 (01)". 
or A— xX = ‘99° = 99 5 
23 23 
~ 499 ~ 99° 


And generally, if C = any circulate containing c digits, we shall 
have 
A— X= C(‘1)'+ C(1)*4+ C(1)*+,,, [2]; 
€ (*1)° C 
~T=(lY (10%—1~999.,.cey Why? Rule? 

We observe that a quantity may increase constantly by an inf- 
nite number of additions, and yet never exceed a determinate finite 
quantity or limit; as ‘333, é&c. constantly approaches to 2 which it 
can never surpass, This gives us our first notion of limits. 

Def. “When a variable magnitude A —_X can be made to (64) 
approach another A which is fixed in such a way as to render their 
difference X lessthan any given magnitude, without, however, their 
being able ever to become rigorously equal, the second A is called 
a limit of the first 4 — X,’* 


A direct consequence of (63) and further illustrating (56), is the 
following important principle : 


* Franceur, Mathematiques pures, p. 166. 


EXERCISES, 59 


Cor. If an equation exist between terms affected by whole (65) 
additive powers of a variable quantity, z, capable of indefinite dim- 
inution, then the constant coéfficients of the like powers of z, will 
be respectively equal to each other. 

Thus, if 4+ A,-2'+A,- 7*?+A,¢ r3+... 

=a-+a,er'+a,-r*+a,+7°+.,.., 
where A, <A,,* Ag, As, ...5 Gy Ay) Ay Az,..., are constant and inde- 
pendent of z which is variable, then 

(63) A=a,and A,+ 7+ A,-z*?+...=a,¢ ¢+a,¢7*+..., 
or A,+ A. r+A,¢27?+...=a,+a,° r7t+a3¢ r?+...3 
and .*. (63) A,=a,, &c., &c., &c. ; which was to be proved. 

If, in a problem, a, a, da ..., can by any means be found, A, 
A,, A,,..., will become known, Applications will be given further 
on. 


EXERCISES. 


I. In Simple Equations. 


1°. Given x+3=5, to find xz. Subtracting 3 from both sides, 
we have 
z2=5—3=2, the Ans. 
2°. Given r—5=7, to find z. 


Adding 5, z=7+5=12, Ans. 
3°. Given 27 =15+-7; 4°, Given 27 —25 = 3x — 75; 
af 22 —xr=15, ’. 1) —25 =3r —2z, 
or, found, eee 15. or 32 — 27 = "75 — 25, 
or x = 50, found. 
5°, Given r—a=), 6°. Given 47+ m= 32r-+n, 
found z=what? found x= what? 


* Read A sub-one, A sub-two, &c., or simply A first, A second, &c. The ad- 
vantage of this notation is that it points out the power of x to which the co- 
efficient belongs; thus, An would belong to 2. 

t For, denoting by A, the greatest of the coéfficients A, Ag, As, ..., we have 

Al + © Ag 2? + Azo 2+... < Age a+ Aye 2? +A, . 23+... 
= Ay. e(1-+a-+2?+...-2") 
1—z . x” 

(49) —s Ay oe ue; (ete? ae 
which diminishes without limit as z approaches zero—and the same may be 
affirmed of 


? 


Yetta. x+a3.23-+.... 


60 EXERCISES, 
7°, Given (m+1)z—p=mz-+4q,|{ 11°. 
found z=p+g. found 
9° O 

8°. Given Dey _ ee Tl ao 
7 7 found 


found z=? 
9°, Given 27 = 8, 


Given 4z = 3. 
© = 4; 

Given mz =n, 
nN 

Tres 

m 


13. Given ax=ab, 


found gus A, found o='b. 
10°. Given 3z = 9, 14. Given az = ab — ac, 
found ve 3. found m =s.t 

15°. Given a a 

found Epa ee ears ee 
g wel) mewnem 
16°. Given — =<, pit re mia 
7 t i t. 
found gi A A ee 
t Ly t t 
17°. Given ar+ br =c, 18°, ax+ bh? =a? + bz, 
(a+ b)t=Cc, ‘ c= a+b, 
ane 
atb 
) re 20°, “ = 

19°, —=a+b, in a a+b-+c¢, 

*. c=m(at+b)=mat+mbd. zZ=(m-+n)(a+b-+c) 
= ma+t+mb+ mc 
+na+nb-+ ne. 

E 
21°, ——___-+= b ; Js ew aie see ore 
Tay Ni a+b+c¢ 22 apa AEN a+b+c—d, 
v= (a+b)? —c?, (a+b)? — (e-d)? 
P 
Posi! TAG Er 
23°, ahaa in Zak Si aoe loc SO te tl A 
p+ 2qr—(¢? +17?) =(p+q—r) 2, 
T=p— qt. 

240, 13} — > = 22 — 84, “5 ae, 

25°, 22+ 7-+ sr = 6r — 23, .«. r = 12. 

26°. 12h+ 32-6 — = _ 53, xz = 1394 


EXERCISES, 61 


wih beekeaBldm tuatitian 


1°. Given 2x-+3y=14 ” tery “ (4—-$2=14-3, 


5r+4y=21 &r+-y=5t 
‘ _4-—4_ 15-8 | E: 
a tiga Se wy! NE, 
2°, 2+22%y4+32=14, A OF) shi 
44 By 1 62'=92, hs ene agi 
a=? yah eat ante eee ’ 
£4 gil ay ee 
4°, ax-+by=c, 5°, axr—by= be, 
ar+by=c; pr — qy = QT 3 
gely=t aS feat ys 


Ill. In Quadratics. 


1°, Given x? +42 = 45, to find z. , 

Adding 4, r?t+4r7+4=49, . 
or, (r+2)? = 49, 

v. g+2= 4%, 

oe g=—2+7=5, or=— 9. 

Pa 37° +6=27°424, 2 =9. 

3°. az? —b=cr?+d,..r=! 

4°, 7— 7? —82? = 6774+ 3-21,..7=? 

a; cnn pam EN: 

a b 


6°. (ea)? +2n(e— 4) =b, . p= a—nt (640°)? 
7°, o@41 re" =a, or y? + Wmy =a, putting y= 2", 2m=T; 


4 i now acre Be 
vs y=—m+(a+m’)’, e=ys=[—m+(e+5) | 2 


8°. x’ —6r? = 16, .. r=? 
10°, z?—2az =2ab4+ 07, .. c=? 
Li?: ru? —2rmu = Wrnut p?r — Wrpm — arp, .. = 2 


1 z$+364 234+ 67 — 182? =0,.. 7 =! 


62 EXERCISES, 


IV. Eliminations producing Quadratics. 


1s ry = 750 cy = a x? + y? = 13001 
— S le Se oe Sts, hig ths te 
— = 3h. —=b6, x? — y* = 1449, 
y 
2 2 — ee ee 
> egal CO Pg) bo.) Nh | 
4 x? + y? = 324900, x?-+-y?=b, 


a 
ry =a = ry=a 
7°. S$ a2z=5 le Meaceeiag | 9°, (yz=b 
y 


: r 
y+2z=c, adie 
rent | 
ee 
bg 
x+y =500 r+ y = 2000 =A] 
10°. {450-2 22 | 11°.) 1710—-2_17e $ 12.3 eh on 
450—y By’ 1040—y 12y'" sagt 
ED eee pore | chai (10°) 9 
LL ytb. 24 y?=514, 
x+y=b. § ary (a—z) (b—-y)=p 
160.) Tiy=min Wa, SS ets 
(a—zx)* + (b—-y)*=s. w+y +z t+ y= 330. 


ES 
1: 
} 
la 
} 
: 
} 


by ‘ e+ ry + ry? = 126 
os gc? —y? = zy. ee ee ee 
208 | 2 29) Cal sae Dh cote} mh ed 
(r+ y) (#* + y*) = 580 r?-+4y?—(x-+y)=42, 
90. r+y+ry=a 930, r+y=a 
z+ y?—(c+y)=b., ides 
24°, r+y=a 25°, e+y+r+y=a 
ry (2? +9’) =b, : shld iadtaak 2 


BOOK SECOND. 


PLANE GEOMETRY DEPENDING ON THE RIGHT LINE. 


SECTION FIRST. 


Comparison of Angles, 


Def. 1. The doctrines of extension constitute the science of Ge- 
ometry. 

Def. 2, Solids have three dimensions, length, breadth, and thick- 
ness, 

Def. 3. The boundaries of a solid are surfaces, the perimeter of 
a surface are lines, and the extremities of a line are points. 


PROPOSITION I. [primary norTion]. 


A Straight Line is such that it does not change its direc- (66) 
tion at any potnt in its whole extent. 

This truth is not produced as a theorem, for it is incapable of de- 
monstration ; not as a problem, for there is nothing to be done; 
neither as a corollary, for it is the consequence of nothing ; nor as 
an axiom, for it is hardly of the unconditional and absolute charac- 
ter of that enounced in the words, ‘* the whole is equal to the sum 
of all its parts ;”’ and it is not a definition, for we gain no new idea 
by the mere terms of the proposition : we only recognize by and in 
them one of those primary notions which we possess anterior to 
all instruction, and which, as they are necessary to, lie at the foun- 
dation of, every logical deduction. It would doubtless be out of 
place to enter here into any investigation in regard to the origin of 
our ideas; but I think it will be apparent, that the notion of con- 
tinuity is codriginal with that of personal identity, and therefore, 
antecedent to argumentation; and continuity measured out on 


64 RIGHT LINE. 


the one hand, in the lapse of events, as the periodic return of day 
and night, the revolutions of the celestial sphere, or the index of 
the chronometer, becomes time, and, on the other, attached to form 
by a personal passage over the surfaces of bodies, it becomes space ; 
and, considered in regard to space and restricted by the notion of 
perfect sameness, continuity gives us the idea of direction, or that 
of the straight line. 

Corollary 1. Two straight lines cannot intersect in more (67) 
points than one; for having crossed once, it is obvious from (66), 
that, in order to a second intersection, one of the lines, at least, must 
change its direction, 

Cor. 2. Straight lines coinciding in two points, coincide (68) 
throughout, otherwise two straight lines would intersect in more 
points than one, which is contradictory to (67). 

Cor. 3. Straight lines coinciding in part, coincide through- (69) 
out, and form one and the same straight line (68). (Why 2) 

Cor. 4. Two straight lines cannot include a space (68). (70) 
(Why 2) 

Def. 4. A Plane is a surface with which a straight line may be 
made to coincide in any direction. 

Def. 5. Two straight lines are said to be parallel when, situated 
in the same plane, they do not meet how far soever they may be 
produced. 

Cor. 5. Through the same point, only one sfraight line can (71) 
be drawn parallel to a given line ;* for the directions of all the lines 
save one, drawn through the given point, will evidently (66) be 
such as to cause them to meet the given line if sufficiently produced. 

Scholium. The last corollary, though commonly given as an ax- 
iom, has been thought not sufficiently evident of itself. 

It is not self evident, doubtless, if regarded as a consequence of 
any mere definition that can be given of a straight line, but neces- 
sarily follows from that idea of a straight line, viz., continuity in 
sameness of direction, which we possess anterior to all definition, 
As such, it is, I believe, as well established as the primary truths in 
any department of human knowledge. 


Application 1. To make a straightedge. Hav- _ ,--_____ 
ing formed a ruler as straight as possible and drawn (@——=———_y 
a line with it upon a plane surface, turn the ruler Hiatt 


ooo 


* When the wu.d dine is used alone, straight line is to be understood. 


ADJACENT ANGLES, 65 


over upon the opposite side, and, with the same edge, repeat the 
line, coinciding with, or better, very nearly coinciding with the 
first; if the lines coincide, or appear equally distant throughout, 
the edge is sensibly straight. 

Application 2. To test the parallelism of the edges of the ruler, 
place it against a second straightedge, and, having drawn a line, 
turn it end for end and draw a second, (Why 2) 

Application 3. To test a plane, apply the straightedge to it in 
different directions. 

Def. 6. When two lines intersect each other, or would intersect 
if sufficiently produced, the inelination of the one line to the 
other is called an Angle. It is obvious that angles are of different 
and comparable magnitudes, and, therefore, like other geometrical 
quantities, solids, surfaces, and lines, are capable of addition, sub- 
traction, multiplication, and division, and, consequently, subject to 
mathematical investigation. 


PROPOSITION II. 


The sum of the adjacent angles formed by one line meet- (‘T2) 
ing another, is always the same constant quantity. 


Let the line AO meet the line BOC in O, ma- A 


king the angle AOB any whatever, and, in like / 
B c 


manner, let the line A’O meet the line B’'OC’ in 


O, making the angle A’OB’ any whatever. Pla- AY N’ 
cing the first figure upon the second, let the point 3’ — C’ 
O coincide with the point O and the line OB take Fig. 3. 


the direction OB’ then will OC take the direction OC' (69) and the 
line BOC will coincide with BOC, also OA will take a certain di- 
rection OA", which we are at liberty to suppose between OA’ and 
OC’; whence the angle AOB will be equal to the angle A”OB,, the 
two becoming identical, and AOC to A”OC’ ; but the angle A” OB’ 
is, by hypothesis, the sum of the angles A’OB’ and A’'OA”, and 
therefore greater than A’OB' by A’OA" (1,), while the angle A” OC’ 
is less than A’OC’ by the same quantity ; therefore the sum of the 
angles A’OB, A”OC’ is equal to the sum of the angles A’OB, 
A‘OC' (1,,), whence the sum of the angles AOB, AOC, is equal to 
the sum of the angles A’OB’, A’OC’ (1,) which was to be proved. 

The same in the language of symbols ; 

Let BOC coincide with B‘OC’,, and OA take the direction OA" ; 

5) 


66 RIGHT ANGLES, 


Z AOB= A'OB = A'OB' + A'OA" 
and Z AOC= A"0OC' = A'0C — A'OA" ; 
Z AOB+ AOC = A‘OB'+A'0OC.. Q. E. D.* 

The method employed in the preceding demonstration is obvi- 
ously that of superposition, and the principles upon which it is 
grounded are the nature of the straight line, the whole is equal to 
the sum of its parts, and, equals added to equals, the sums are 
equal, 

Def. 7. If the angle AOB= AOC, then AOB and 
AOC are called Right Angles ; hence, AOB or AOC ri 
being the half of AOB-+ AOC = a constant quantity 
(72), is also a constant quantity, or, 

Cor. 1. All right angles are equal to each other. (73) 

Cor. 2. The sum of the adjacent angles formed by one (74) 
line meeting another is equivalent to two right angles. 

Cor. 3. Conversely ; two lines met by a third, so as to (75) 
make the sum of the adjacent angles equal to two right angles, 
form one and the same straight line. 

For if not, let BOX be a straight line, while 

Z AOB+ AOC =+—,,t [by hypothesis; ] 
then (74)  AOB+AOX=-; Beettaey 

.. subtracting AOX — AOC = 0, Fig. 33. 
therefore BOC does not differ from the straight line BOX, 

Cor. 4. The vertical angles formed by intersecting lines (76) 
are equal. 


If a@ and 6 be vertical angles, while c is adjacent to <a 
z 


Fig. 32. 


both, we have (74) a+c=-, 

and b+c=4+; Fig. 34. 
ase at+tc=b+ce, 

and (1,,) a=); 


Cor. 5. The sum of all the angles that can be formed ata (77) 
given point and on the same side of a straight line, is equal NV 


to two right angles (2). Fig. 35. 

Cor. 6. The sum of all the angles that can be formed (78) 
round a given point, is equivalent to four right angles. aie 

Ape? 

Fig. 36. 


* “Quod erat demonstrandum,” which was to be proved. 
t We shall use the symbol —+- for two right angles, and +- for four, while - will sig- 
nify a single right angle. 


SURVEYOR’S CROSS. 67 


Application 1, To make a Rightangle for drawing perpen- 


diculars. Form a triangle, having one angle as y 
nearly right as possible ; set its base upon a straight Al 
edge and draw a line by its perpendicular, turn the Hii 
triangle over upon the opposite face, and repeat the Ji 


line. The instrument may be three inches in base *—______? 
and six in altitude, and cut from pasteboard, brass, 
or ivory, or framed of wood. 

Application 2. To make a right angle in the field. Take a 
piece of board about 10 inches square, bore a hole through the 
centre, and fit to it a staff 44 feet long, so that, when the staff is 
stuck vertically in the ground, the board shall turn freely upon its 
top and continue during a complete revolution in the same plane, 
which will be determined by sighting its upper surface at a given 
object; draw two lines at right angles to each other through the 
centre and from corner to corner. In these lines at the corners 
may be stuck four needles—and to know whether they are accu- 
rately at right angles, we have only to place the board as nearly as 
possible in a horizontal position, and then to sight at two marks in 
range with the needles, in the lines drawn at right angles, and to 
see if we hit the same marks when the board is turned a mht 
round, Why? 

The Surveyvor’s Cross is the same instrument, 
only better finished. 

The sight vanes (fig. 5,) are hairs opposite slits, 
and vice versd; and the piece that turns upon the 
staff is represented in fig. 5. Fig. Bs 

To survey with the cross, it is necessary to be Int aij 
provided with two straight flag-staffs, which should Fig. 53. 
be wound with a red flag; or they may be square, and the adjoin- 
ing faces painted red and white. In running a line, one staff is to 
go before and the other is to be left at the last position of the cross 
for a back sight. A measuring rod, tape line, or Gunter’s chain 
is also to be provided, The chain is 4 rods or 66 feet in length, 
and centesimally divided by a hundred links, each, consequently, 
equal to 7:92 inches, It is a maxim in surveying land that all in- 
struments, whether for measuring lines or angles, must be kept in 
a horizontal position; for it is the base, or the projection of the 
field upon the same horizontal plane that is required. 


68 PARALLEL LINES. 


Let it be required to survey the field ABCDE. Boog 
We set the cross at B’ in the line AE so as to make 
BB’A a right angle, and in the same way we find 4 3B Cc’ DE 
the points C’, D’; we then measure the lines AB’, Fig. 6. 
B'C’, C’'D’, D’E, and the offsets B'B, C’C, D D, when the field is 
surveyed. This is a good method of determining x c 
an irregular boundary, ABCDH, as that of a river. / 
Frequently it will be better to take the offsets from A D 
a diagonal, AC, Fig. 6. 
Sometimes it will be preferable, especially if the corners cannot 
be seen from each other, to measure AB then BB’ 8. © 
at right angles to BA, then BC at right angles to Pa 4 
B'B, then CC’ at right angles to CB’, and finally A D 
C'D at right angles to C’'C. The student should Fig. 63. 
now make an actual survey and draw an accurate plot of it on pa- 
per. This may be done by aid of the ruler, rightangle, and a Jine 
of equal parts, which are to be run ; ; ' ; . H . i + ri +t 
off of any convenient magnitude by Fig. 7. 
a pair of dividers. 


PROPOSITION III. 


Two lines intersected by a third, making the alternate (79) 
angles equal to each other, are parallel. 


Let the lines AB, CD be inter- 
sected by EF in O and O’, making L 
Z AOO' = 00D; then will AB, 
CD, be parallel. Ifnot, let AB, CD, 
produced, meet on the right in I, Fig. 8. 
then it may be shown that they will meet on the left in a point I’, 
and consequently that two straight lines, IBAI’, IDCI’, may inter- 
sect in two points, which is contrary to (67). | 


For, since Z BOO'+ AOO'=+=C0O'O0+ DOO, 
and by hypothesis, Z AOO'= DOO, 
.. subtracting, Z BOO’ =CO'O; 


from which it follows that the figure BOO'D may be reversed and 
applied to the figure CO'OA; for the line OO’ being reversed and 
applied in OO, the line O'D will take the direction OA and there- 
fore coincide with OA through its whole extent, and, for a like 
reason, OB will coincide with O'C. 


PARALLEL LINES. 69 


Hence, if the lines OB, O'D, intersected in I, the lines OA, 
O'C, now coinciding with them in their reversed position, would 
have the same point of intersection in I’. 


Cor. 1. Conversely, two parallel lines intersected by a (80) 
third, make the alternate angles equal. 

For, if the lines AB, CD, being parallel, do not make [ fig. 8.] 
the angles AOO’, OO'D equal, draw A’OB' (the student will draw 
the line) making 7 A'OO' = OOD, then will (79) A’B’ be parallel 
to CD, but, by hypothesis, AB is parallel to CD, .°, through the 
same point O two lines AB, A'B’, have been drawn parallel to the 
same line CD, which is contrary to (71). 


Cor. 2. The equality of two alternate angles determines: (81) 

1°. The equality of all the other alternate angles, whether in- 
ternal or external. (Where is this shown 2) 

2°. The equality of the external to the opposite mternal on the 
same side of the secant line. (Why ?) 

3°. The sum of the internal angles, or the sum of the external, 
on the same side of the secant line, to be equal to two right angles, 

For, if the angle DO'E= AOF, adding 7 BOF to both sides, we 
have 

Z DO'E+ BOF = AOF + BOF =~. 

Cor. 3. If the secant line be perpendicular to one of two (82) 

parallel lines, it will be perpendicular to the other also. 


Cor. 4, Two lines parallel to the same line, are parallel to (83) 
each other. [How can this be shown from (82) ?] 

Cor. 5. Two angles having their sides parallel, and lying (84) 
in the same direction, are equal. For, producing the /£ /e_ 
sides of the angles a, 5, till they meet, forming the an- | 
gle c, we have (81, 2°), a=c=b. g- 


Fig. 82. 
Cor. 6. The opposite angles of a Parallelogram are equal. (85) 
It is scarcely necessary to remark that a parallelo- ; 

gram is a quadrilateral, or four-sided figure, having its BES: 

: ‘ Fig. 83, 

opposite sides parallel. 

Application. Draw parallel lines upon paper with 
the straightedge and rightangle, and construct them 
in the field by aid of the cross, 


Fig. 8. 


70 ANGLES OF POLYGONS. 


PROPOSITION IV. 


The sum of the external angles of a Polygon, formed by (86) 
producing the sides outward, is equal to four right angles. 

Let a, 5, c, d, e, be the external angles of a 
polygon; through any point, either within or out 
of the polygon, draw lines parallel to the sides of 
the polygon, forming the angles a’, b’, c’, d, e’, 
corresponding severally to the angles a, d, c, d, 
e; then (84) we have 

La=a,b=b, ca'/d dyer Fig. 9. 
, adding, a+b+c+d+e=a4+b)4+c4d +e =4+. QED. 

If we denote the corresponding internal angles by ‘a 
A, B, C,..., of which we will suppose there are n, of regen: 
we shall have (74), 

A+a=+, B+b=+, C+ce=4, ... [n] 
A+B+C+..4¢a+6+ce-4... =n (-), 
but (86) a+b+e+...=2(+); Fig. 92. 

.. subtracting, A+ B4+C-+...[n]=(n— 2) (+). 

Cor. 1. The sum of the angles of a polygon is equal to (87) 
two right angles taken as many times as there are sides less two. 

Making successively n = 6, 5, 4, 3, we have 

(n — 2) (+) =4 (+), 3(+), 2(+), (+); therefore 

Cor, 2. The sum of the angles of a Hezagon is equal to (88) 
eight right angles, 

Cor. 3. The sum of the angles of a Pentagon is equal to (89) 
the sum of six right angles. 

Cor. 4. The sum of the angles of a Quadrilateral is equiv- (90) 
alent to four right angles, 

Cor. 5. All the angles of a quadrilateral may be right an- (91) 
gles; such a figure is called a Rectangle. 

Cor. 6. The sum of the angles of a Triangle is equal to (92) 
two right angles. 

Cor. 7. The sum of the acute angles of a Right angled (93) 
Triangle is equal to a right angle 

Cor. 8. The external angle formed by producing one of (94) 
the sides ofa triangle, is equal to the sum of the two Opposite in- 


ternal angles, Let a, b, c, be the three angles of any tri- ¢ 
angle, and ¢ adjacent to the external angled; we have 


ad 
Fig. 93. 


EXERCISES. 71 


Za+b+c=4, 
and (?) d+c=-; 
(2) Ld=a+b. 


Scholium. The angles of a triangle will be determined : 

1°. When the three angles are equal ; (how 2) 

2°. When two angles are equal and the third known; (how 2) 

3°. When two angles are given. (How ?) 

What is a hexagon? pentagon? quadrilateral? triangle? right 
angled triangle ? 


EXERCISES, 


1°. By aid of the straightedge and rightangle, construct 
about the vertex of a triangle angles equal to those at the 
base, and do this for triangles of different forms. 

2°. Through the vertex of any triangle draw a line paral- 
lel to the base and prove (92). 

3°, The same by drawing a parallel to one of the oblique sides. 

4°, The same by drawing perpendiculars to the base. 

5°, From any angle of a polygon, draw diagonals to all the other 
angles and prove (87). 

6°. From any point within a polygon, draw lines to the several 
angles and prove the same. 

7°. Let fall a perpendicular from the right angle of a WA 
right angled triangle upon the hypothenuse, and prove 
that the three triangles thus formed are equiangular, 

8°. Having drawn the sides of two angles respectively | ae 
perpendicular to each other, prove that the angles are \ 
either equal or that one is what the other wants of two Fig. 12. 


Fig. 10. 


Fig. 11. 


right angles. | 
9°. Prove that two triangles are mutually equiangular 
when the sides of the one are severally perpendicular to the 


sides of the other. Fig. 13. 


SECTION SECOND. 
Equal Polygons.—First Relations of Lines and Angles. 


PROPOSITION I. 


If, in two Polygons, excepting three parts which are (95) 
adjacent, viz., two angles and an included side, or two sides and 
an included angle, the remaining parts, taken in the same order, 
are severally equal, the Polygons will be equal throughout, and 
the excepted parts will be equal, each to each. 


First. Let AB = A'B,, 7 B=B,BC=BC, B C 


ZiCr%s.C' se aig’ ; then, applying the figure ni Pret 
ABCD to A'BCD,, let the point A fall on A’ and B ) 


the line AB take the direction A'B’; the point B ‘ \ 
will fall on B’, since AB=A'B’ by hypothesis, and “~/ eo isd 
BC will take the direction B'C’, because the angle ah 

B = By, by hypothesis, and the point C will fall on C’; (why 2) 
and finally, the line CD will coincide with C’D’, (why?) so that 
the points A and D coinciding with A’ and D’, the line AD will 
coincide with A’D’, and the polygons will coincide throughout, and 
ZAwill= A’, AD=A'D,, D=D". 

Second. Let 7A=A', AB=A'B, 2 B=B, BC=BC, ZC 
-=C’'; then by a superposition altogether similar to that above, it 
may be made to appear that the polygons will coincide throughout, 
and that the three adjacent parts, CD, Z D, DA, will be severally 
equal to the three, C’D’, 7 D’, D’A’; and itis obvious that the same 
reasoning may be extended to polygons of any number of sides. 
Q. E. D. 

Cor. 1. If two triangles have two sides and the included (96) 
angle of the one, equal to the two sides and included angle of the 
other, each to each, the triangles will be equal, and the remaining 
three parts of the one respectively equal to the remaining three 
parts of the other. ws 

[The student should letter the figure and explain. ] Fig.149 

Cor. 2. If two triangles have two angles and the included (97) 
side of the one, severally equal to the two angles and the included _ 


PARALLELOGRAMS, 73 


side of the other, the triangles will be equal throughout. [By 
what figure, and how, may this be illustrated ?] 
Cor. 3. The diagonal divides the parallelogram into two (98) 
equal triangles; .°. = | 
1°. The opposite sides of aparallelogram are equal, | : 
2°. Parallels are everywhere equally distant, Fig. 143. 
{Letter the figure and explain carefully. What condition must 
be imposed upon the parallelogram in order to illustrate the last 
part of the corollary 2] 
Cor. 4, The diagonals of a parallelogram mutually bisect, (99) 


or divide each other into equal parts (98, 1°), (80), ~s *>7 
hi aa oil 


(97). 
¥ Fig. 144. 
Cor. 5. An Isosceles triangle is bisected by the line (100) 
which bisects the angle embraced by the equal sides. Cc 
For let CA = CB and 7 ACD =BCD; Q . 
then (96) will A ACD = BCD; .-. ee 
Fig. 145, 


Cor. 6. In an isosceles triangle, the angles opposite the (101) 
equal sides are themselves equal; as 7 A = B, and 

Cor. 7. The line bisecting the vertical angle of an isos- (102) 
celes triangle bisects the base, to which it is also at right angles or 
perpendicular, 

Cor. 8. Equilateral triangles are equiangular (100). (103) 


PROPOSITION II. 


In any triangle, that angle is the greater which is op- (104) 
posite the greater side. 
Let CB be > CA; then will 7 CAB > CBA. } 
For, taking CD, a part of CB, equal to CA and join- 
ing AD, we have ys Be 
ZCAB>CAD=CDA=ABD-+BAD>ABD. “* ve 


Q. E. D. 
[Give the reasons, Consult (?), (101). (94), (?).] Fig. 15. 


Cor. 1. Conversely, the side opposite the greater angle, (105) 
is the greater, For if the side opposite the greater angle be not 
the greater, it must be equal or less; it cannot be equal, for 
then would the angles be equal (?), neither can it be less, for 
then would the opposite angle be the less (104), which is contrary to 
the hypothesis. 


74 SHORTEST DISTANCE. 


Cor. 2. A triangle having two equal angles, is isosceles (106) 
(105). 

Cor. 3. An equiangular triangle is equilateral (106). (107) 

What is an Isosceles triangle? What an Equilateral triangle ? 


PROPOSITION III. \ 


Any two sides of a triangle are together greater than (108) 
the third. 


Let ABC be any triangle; then will the sum of /D 
any two sides, as AC-+ CB, be > the third AB, a 


Produce AC to D, making CD = CB, and join ae 
DB. ’ 
A> 


then (101) ZCDB = CBD < ABD; B 
“ (105) ABK< AD=AC+CD =AC-+CB. Fig. 16. 
Q. E. D. . 


Cor. 1. The sum of the lines joining any point within a (109) 
triangle and the extremities of one of its sides, is less than the 
sum of the other two sides. 


C 
Let D be any point in the triangle ABC ; produce 
AD -to méeet'CBin Fothen. ©) 3 ' “© Weenie AE 
AD+DB< AD+ DE+ EB< AC+ CE-+ EB. : 
(Why?) Q. E. D. +. B 
Fig. 16. 


Cor. 2. Any side of a polygon is less than the sum of all (110) 
the other sides. 
[Letter and show how. ] 


Fig. 163. 


. Cor. 3. Of two polygons, standing upon the same (111) 
base and one embracing the other, the perimeter of the enveloping 
is greater than that of the enveloped. 

[See the last figure and explain. ] 

Cor. 4. The straight line is the shortest that can be (112) 
drawn from one point to another, 

First. Let APB be any line, whether curved 
or broken, but concave toward the straight line 
AB, then APB will be greater than AB. For, 
taking any point P, in the line APB, and revoly- Fig. 164. 


SHORTEST DISTANCE. 75 


ing the branches AP, BP, about A and B till the point P coincides 
with the straight line AB, first in Q, then in R, the straight lines 
AP, BP, in the positions AQ, BR, will overlap each other by a 
certain line RQ, since the st. line AP + st. 1. BP.> AB—and there- 
fore, the branches AP, BP, falling on the same side of AB, since, 
by hypothesis, the path APB is concave toward AB, will intersect 
each other in some point E; whence the path APB, being equal 
to the sum of the paths AE+ EQ and RE-+ EB, will be longer 
than the path AEB by the sum of the parts RE and QE. The 
same may be shown of any other path that is concave toward AB 


except AB. 

Second. Let the path AcdefghB be any what- eee 
ever, having flexures either continuous or abrupt Zz, on 
in c, d, e, f, g, h; then drawing the straight lines A B 
Ac, cd, de, ef, fg, zh, hB, Fig. 16s. 


we have (110) AB < polygonal perimeter (Ac + cd-+de+ef 
+fe+geh+hB) < the path (AcdefghB), 
as shown aboye, since all the parts of AcdefghB are concave to- 
ward the several sides of the polygon. Q. E. D. 
Cor. 5. ‘* Hence also, we may infer, that of any two (113) 


paths, ACB, ADB, leading from A to B, and every- c 
where concave toward the straight line AB, that “1-2 — 
which is enveloped by the other, as ADB, is the 4 B 


shorter. For of all the paths not lying between Fig. 166. 
ADB and the straight line AB, there is none, ADB excepted, than 
which a shorter may not be found. And this is the case whether 
the paths ACB, ADB, be both of them curvilineal, or one of them, 
(ACB or ADB), rectilineal.”’ * 

Cor. 6. The perpendicular is the shortest distance from (114) 
a point to a straight line; and of oblique lines, that is the shorter 
which is nearer the perpendicular—and those equally distant are 
equal. 


For Jet PA be perpendicular to CBAB, and AB’ P 
= AB < AC, and produce PA in Q, making AQ = AP; eri 
joining QB, QB, QC, we have Cc BA B 
PC+QC> PB+QB> PA+QA_ (2) b 
.. dividing by 2 PC>PB> PA, (2) Fig. 167. 
also PR= PH. (2) 


* Library of Useful Knowledge, Geometry. - 


716 ANGLES AND SIDES. 


Cor. 7. All points equally distant from the extremities (115) 
of a straight line are situated in the same perpendicular passing 
through the middle of that line. 

The points C, B, A, B’, are equally distant from P and Q. 


PROPOSITION IV. 


If two triangles have two sides of the one equal to two (116) 
sides of the other, each to each, but the included angles unequal, 
then the third sides will be unequal, and the greater side will be 
that opposite the greater angle. 


Let two of the equal sides be made common 
in AB, and the other two, AC, AD lie on oppo- 
site sides of AB; draw AI bisecting the angle 
CAD and terminating in I, which will be a point 
in CB on the hypothesis that CAB is the great- 17. 
er angle, since then we have 7 CAB> CAI= DAL S DAB; — 
finally, join ID. We have (96), (108), 

BC=BI+I1IC=BI+I1D>BD. Q@. E. D. 
Cor. 1. Conversely, if two triangles have two sides of (117) 
the one equal to two sides of the other, each to each, but the 
third sides unequal, the opposite angles will be unequal and that 
will be the greater which is opposite the greater side. For it can 
be neither equal nor less. (Why 2) 


q 


Cor. 2. If two triangles have the sides of the one seve- (118) 
rally equal to the sides of the other, the angles opposite the equal 
sides will be severally equal. 

For, if AB= A'B, BC=BC, AC=AC'; C old 
*- ZC cannot be 2C, for then would AB2A'B ; 

-- ZO=C, and, for a like reason, 7 A =A’, <3 4G’ 
B=B. Fig. 172. 

Cor. 3. When the opposite sides of a quadrilateral are (119) 

equal, each to each, the figure will be a parallelogram [ fig. 14,]. 


EXERCISES. v iy j 


EXERCISES, 


1°, It is a principle in Optics that a ray of 
light, AP, impinging upon a reflecting surface, 
MM’, makes the angle of reflection BPP’, equal 
to the angle of incidence APP’, PP’ being per- 
pendicular to MM’ at P. Given in position the 
luminous point A, the eye, B, and the mirror, Fig. 18. 
MM’, to find the point P from which the light is reflected to B, and 
to prove that APB is the shortest path from A to B by way of 
the mirror, or that AP-+ PB < AQ+ QB. 

2°. Having given in position the elastic planes A 
PQ, QR, and the points A, B; itis required to find 
the track of an elastic ball projected from A upon er R 


PQ and rebounding from QR so as to hit a pin 
DAL Fig. 182 


td 


3°. Given in position the straight line MN and the jf \ 
points A, B; required the point O in aN such that 
AO shall = BO. WY 


4°. Prove that the perpendiculars drawn through mS MIDS 
the middle points of the three sides of a triangle, will LOX 


intersect in the same point. 


5°. Having given in position the lines AB, CD, ; 
and the point P,it is required to draw through Pa i 


line mn, which shall make equal angles with AB, “2—— x 
B 
OM Fig. 21 


6°. Let AOB be a right angle, COB an equilateral - c 
triangle, and OP perpendicular to CB. Prove that 
the angles AOC, COP, POB are all equal, How e 


could you trisect a right angle? K \. 


Fig. 22. 
7°. Let A, B, two angles of an equilateral triangle, be bisected 
by the lines AO, BO, and from O draw OP, OQ, parallel to the 
sides CA, CB, and terminating in the base in P and Q. Prove 
that AP = PQ=QB. [The student will construct the figure. ] 


78 PROPORTIONAL LINES, 


8°. Prove that any side of a triangle is greater than the differ- 
ence of the other two. 

9°, Prove that the angle included between two lines drawn from 
the vertex of any triangle, the one bisecting the vertical angle and 
the other perpendicular to the base, is half the difference of the 
basic angles. 

10°, Prove that if a line be drawn from one of the equal angles 
of an isosceles triangle upon the opposite side and equal to it, the 
angle embraced by this line and the production of the base, or un- 
equal side, will be three times one of the equal angles of the 
triangle. 

11°. Draw three lines from the acute angles of a right angled 
triangle—two bisecting these angles and the third a perpendicular 
to one of the bisecting lines—and prove that the triangle embraced 
by these lines will be isosceles. 


SECTION THIRD. 
Proportional Lines. 


PROPOSITION I. 


The segments of lines intercepted by parallels are pro- (120) 
portional. 

In the first place, let. the secant 
line AC be divided into commensu- 
rable parts by the parallels AA’, 
BB’, CC, for instance, such that AB 
containing three measures, BC shall 
contain two of the same; it may be 
shown that any other secant line, 
A'B'C' will be divided in the same ratio. For, dividing AB, BC, 
into 3 and 2 parts in a, 6, c, and drawing aa’, bb’, cc’, through the 
points of division parallel to AA’ and through A’, a’, b’, &e. (points 
of A'C’), drawing A’a”, a'b", &c., parallel to AC and terminating 
severally in the parallels aa’, bb’, &c., we have (?) (2), 


PROPORTIONAL LINES. 79 


A‘a’ =A =eb3rb", 
and /a’A’a =6'ab, Za'aA' =b'd a; 

«. A’a’ =a’b’, and for a like reason, a’b' = b'B’ = Bc’ = c'C’. 

* AB: BO: : 3:3, and A'B’: B@i: 33% 2. 
ow AB?BC:.: AB #aBC’. 

The same reasoning will be applicable, it is obvious, whatever 
may be the number of parts into which AB, BC, may be divided, or 
whenever AB, BC are commensurable. 

Next, let the segments a, b, be incommensurable. 
Produce 6 so that 6+ 2 shall be commensurable 
witn @, and through the extremity of z draw a 
parallel increasing the corresponding segment b’ by 3 
xz; then will 6'’+ 2’ be commensurable with qa’; Fig. 232. 
and, by what has already been demonstrated, we shall have the 
proportion 


b 


ba - Bt 2 (ee HT oh 
Thun ath ake ek al 


but z can be made less than any assignable quantity, since it is less 
than the measure of a, which may be diminished at pleasure; 


Ul 


whence it follows that a? qi are variable quantities capable of 


indefinite diminution, and, consequently (63), that 


vale. past ha a s37@:8. QE. D. 
mee 


Cor. 1. When three lines, two of which are parallel, are (121) 
cut by two others so as to make the segments of the secant lines pro- 
portional, the third line is parallel to the other two. my 
For, if the line CC’ be not parallel to AA’, let CC” 
be so; then there will result (120) 


AB:BO: 2 A'R:? BC’; , 
rs Sree Ws c” 
BC els vat? Fig. 233. 
but, by hyp., AB: BC: : A’B’: BC, or BC = soe be B : 


“. BC’ =BC, which is absurd; therefore CC’ is not otherwise 
than parallel to AA’. 

Cor, 2. The sides of mutually equiangular triangles are (122) 
proportional. 

For, let any two of the equal angles be made vertical, then will 


80 SIMILAR TRIANGLES, 


the bases m, n, be parallel, and, if through thie 
common vertex a line be drawn parallel to m, n, 
dividing their intercepted perpendicular into the 
parts a’, 5”, we Ay find (120) 

a:b: ih eae. DS ie te, 

Remark. The pcb sans sides are opposite equal angles. 

Cor. 3. If two triangles have an angle of the one equal to (123) 
an angle of the other and the sides about the equal angles propor- 
tional, the triangles will be mutually equiangular and consequently 
similar. Draw a line through the extremity of 5 [ fig. 23,], see 
(122), and imitate the reasoning under (121), 

Def. Figures which have their angles respectively equal when 
taken in the same order, and their sides proportional, also taken in 
the same order, are called similar, 

Cor. 4. The altitudes of similar triangles are to each oth- (124) 
er as their bases. For we have (122) 


tt 


oa b 2 des 64° mark 


Cor. 5. If lines be drawn from the vertex of a triangle to (125) 
points in the base, all lines parallel to the base will 
be divided into parts proportional to the segments 
of the base. 

For the segments ab, bc, cd, have to AB, BC, 
CD, the constant ratio of the perpendiculars Op, > Pie: 23 
OP. 


Cor. 6. Triangles having their sides severally parallel are 


12% 


similar, la 


Fig. 236). 
Cor, 7. Triangles which have their sides respectively per- 


pendicular, are similar. Thus, a, d, c, are to each i 
other as a’, 5’, c’, where it is obvious that the homol- b 


ogous sides are those that are perpendicular to each 
other. 


PROPOSITION II. Fig. 23 . 


A right angled triangle, by a perpendicular let fall from (128) 
the right angle, is divided into two partial triangles, similar to 
itself, and consequently to each other. 


HYPOTHENUSE, 81 


Let a, b, denote the sides about the right angle, \ 
h the side opposite, or the hypothenuse, and m, n, b 
the segments of A made by the perpendicular, p. ns 
The angle (a, h) is common to the triangles (a, b, Fig. 24. 


h), (a, m, p), which also have a right angle in each, and are there- 
fore equiangular, (7 (d, h) being = Z (p, a),) and consequently sim- 
ilar. So it may be shown that A (d, n, p) is similar to A (A, b, a), 
and the proposition is demonstrated. 

Cor. 1. The perpendicular is a mean proportional be- (129) 
tween the segments of the hypothenuse. 


For we have ey ee a 


Cor, 2. The sides about the right angle are mean pro- (130) 
portionals between the hypothenuse and the adjacent segments. 
For we have (122) 

h AMP 


a 
— =—, and —=—. 
a m b n 


“Cor. 3. The segments of the hypothenuse are to each (131) 
other as the squares of the sides opposite them ; for, from the above 
we have 


hm = a’, 
and hn = 6°; : 
.*. by div., mwa 5.5%, 


Cor. 4, The square of the hypothenuse is equal to the (132) 
sum of the squares of the other two sides. For we have 

ae (m+ n)h = a? + B?, 
or hh = h? = a? + 32, 

Scholium. This proposition, the celebrated 47th of Euclid, is 
obviously a direct consequence of (122), and with that theorem, 
which embraces the characteristic property of the triangle, consti- 
tutes the working rules of all geometry. 

Cor. 5. Conversely, if the square of one side of a tri- (133) 
angle is equal to the sum of the squares of the other two, the trian- 
gle is right angled. 

For suppose the triangle (h, a, b) to be such that 
h?=a?-+ b?; then if on @ we construct the triangle 
(h', a, b') making b'= 6 and the Z (a, b') = “, there 
will result A’? =a?+ b2=@?4+ B=h’, .. h' =h, and ya e 

. Z(a, b)=(a, b)=+. Fig. 242. 

6 


82 . BASE DIVIDED. 


Cor. 6. The diagonal of a square is incommensurable (184) 


with its side, the former being to the latter as ./2 to 1, 

For we have xz? =a? + a*® = 2a? 
ie w= da/d, or 2: a= fe J. Hae. 
The student might very naturally suppose that a common mea- 

sure to any two lines could be found by taking the measuring unit 

sufficiently small; but, from this corollary, it is shown that the con- 


trary may be true, and we are therefore taught the necessity of 
extending our demonstrations to the case of incommensurability, 


PROPOSITION III. 


To find the relation between the oblique sides, 
a, b, of a triangle, the line c drawn from the 
vertex to the base, and the segments, m, n, of 
the base made by this line. 

Drop the perpendicular p, intercepting the seg- 
ment z between the foot of p and that of c; we 
have (132), (8), (9), 

a? = p? + (m +2)" =p? + m* + Ime + 2%, 
b? = p? + (n — zr)? = p? +n? —2Qnz4+ 22, 
c? =p? +2; 
.. subtracting the 3d from the Ist, we have 
a? —c* = m*+ 2mz, 
from the 2d b?’—c?=n?—2nz; 


... dividing one by m and the other by n, 


=m + 22, 


b? — ¢? 


=n— 27, 


a®*—c*?  b?—¢? 
m n 

If in any triangle, a line be drawn from one of the angles, 

terminating in the opposite side, and the squares of -the sides 

embracing the divided angle be severally diminished by the 

square of the dividing line, then the sum of the quotients arising 

from dividing these remainders by the corresponding subjacent 
segments, will be equal to the divided side. 


=M-+N; 1, Gy (135) 


QUADRILATERALS. 83 


If we take the particular case in which m and n are equal, there 
will result 
a*—c*? b*%—¢? 
m 
whence a? + b? = 2c? + 2m?, or, 

Cor. 1. If a line be drawn from the vertex ofa triangle (136) 
to the middle of the base, the sum of the squares of the oblique 
. sides will be double the sum of the squares of the middle line and 
the half base. 

If the triangle be isosceles, there will result 


=m+m, 


gece? g® ~ ¢2 


peat 


or a? —c? =™mn. 


=m+n, 


Cor. 2. In an isosceles triangle, the square of the oblique (137) 
side diminished by the square of the middle line, is equal to the 
product of the segments of the base. 

Let ABCD be any quadrilateral, M, N, the middle 
points of its diagonals, AC, BD; join AN, NC; 


then (136) 
AB? -++ AD? = 2AN? + 2BN?, 
and CB? -+ CD? = 2CN? + 2BN?; Fig. 252. 
AB? + BC? + CD? + DA? = 4BN? + 2(AN? + CN?) 
= 4BN? + 2(2AM? + 2MN?) 
= (2AM)? + (2BN)? + 4MN? 
=AC? + BD? -+ 4MN?, 


.. Cor. 3. The sum of the squares of the sides of a (138) 
quadrilateral, exceeds the sum of the squares of its diagonals by 
four times the square of the line joining the middle points of the 
diagonals, 

Cor. 4. The sum of the squares of the sides of a parallelo- , (139) 
gram, is equal to the sum of the squares of its diagonals. : 

For (99) the diagonals bisect each other. y 


PROPOSITION IV. 
To find the distance of the foot of the perpendicular of a trian- 
gle from the middle of its base. 


Let a, b, be the oblique sides of a triangle, and p its perpen- 
dicular, dividing the base 2m into two parts, m-+7, m—2Z; 


84 PERPENDICULAR. 


where, consequently, xz is the distance of the foot 
of the perpendicular to the middle point of the 
base. We have (132) 


a* = p?+(m-+ 2)? 


and b? = p?+(m—z)?; 
i a® —b?=4mz, 

or (10), (a+ 5b) (a— b) =2m « 2z; 
s*. (43) 4 2m:a+b=a—b: 2x, 


The base of any triangle is to the sum of the oblique (140) 
sides as their difference to the double distance of the foot of the 
perpendicular from the middle point of the base. 

This theorem is convenient for finding the perpendicular let fall 
upon any side of a triangle; for we have 


p=[a?—(m-+2)"]' =[(a+m+2)(a—m—2)j* (141) 


PROPOSITION V. 


If aline be drawn bisecting the vertical angle of a (142) 
triangle, the segments of the base thus formed, will be to each 
other as the sides opposite them. 


In figure 16, draw CE parallelto DB, E beinga point [fig. 16] 
in AB; then (81), (80), 
Z ACE= CDB = CBD = BCE, 
and (120) ABei EBs AC: CD=CB. Q. E. D. 


EXERCISES, 


1°, Wishing to 
ascertain the dis- 
tance AB of an in- 
accessible object B, 
I measure a line 
AC at right angles 1 
to AB, equal to 12 i Fig. 26. 
chains, then I take CD back and at right angles to AC, equal to 5 
chains ; and find that a line sighted from D to B intersects AC 
at O, distant from C 325 chains. What is the distance from 
A to B? 


EXERCISES. 85 


We have AB: AO=CD:CO, 
o 5 5 «8°75 

or 12 305 ~ 305 °° = 305° = = 13°46. 

How can the same thing be done by (129)? 

2°, What must be the length of a ladder to turn between two 
buildings, one 20, the other 30 feet high, and 40 feet apart; and 
what must be the distance of its foot from the first-mentioned 
building ? 

Let the length of the ladder be denoted by y, and 
the distances of its foot from the buildings by 20+ z aN 
and 20—z; then ‘ 


(20+ ayn + 20? — yf = ae (20 te! x)? cs 30°, ie meas 
or —- A’. 2- Wr+ 2? + 2? = 2’ —2- Wer+2’ Fri 
+ 30°; 
4 + 202 = 30? — 20°’ = (30 ++ 20) (30 — 20), 
50-10 50_ 


and the distance required is 20 + x = 2625 ; 


length of ladder = y = (207+ 26252)" =? 

3°. Wishing to ascertain the distance from A to D (fig. 63), 
rendered inaccessible by the intervention of a ledge of rocks, I 
fetch a compass, ABB'CC’D, viz., AB=11 chs., BB’ =8, BC=8, 
CC’ =4, and C’D=5. Whatis the distance from A to D? 

Ans. 15 chs. 

4°. Given the perpendiculars let fall from A and B_ [jfg. 18] 
upon the plane MM’, the one 7 the other 5 feet, and the distance 
of these perpendiculars = 10 feet; to find the distance of P from 
the greater perpendicular. Ans. 5 feet 10 in. 

5°. Given AB = a and the distances of A and B from [ fig. 19] 
MN, =p, p'; to find AO=zc. 

6°. Given the hypothenuse of a right angled triangle, h, and the 
sum of the sides about the right angle, = 2m ; to find these sides. 

Denote them by m+ 2, m— 72; aa 


* AY 


4 a \ 
then (132) 2=+ (4h? — m?)” ; Fig. 28, 


ae m+ xr2=m = (th? —m’)’, 
and m—xr=m= (4h—m ah ;—where it is ob- J PRE 


vious that both values given by the double sign + Fig. 282. 
equally satisfy the conditions of the problem. 


86 EXERCISES. 


7°. Given the hypothenuse of a right angled triangle and the dif- 
ference of the sides about the right angle, to determine the triangle. 

If the hypothenuse be denoted by 2a and the difference of the 
legs by 2m, the answers will be 


a: 1 
zt m = + (2a? — m’)* + m, x — m = + (2a? — m?)* —m. 


8°. Given the base of a right angled triangle, b, and the sum 
of the hypothenuse and perpendicular 2m, to find these sides. 

9°. Given the base of a right angled triangle and the difference 
of the hypothenuse and perpendicular, to determine the triangle. 

10°. A liberty-pole 100 feet long was broken, and, resting upon 
the stump, its top fell at the distance of 40 feet. Required the 
length of a ladder, planted 10 feet distant, that shall reach the 
break. 

11°. A greyhound is 10 rods distant from a hare; the hare 
starts off at right angles to the line joining them and the hound 
pursues in the hypothenuse of a right angled triangle to intercept her. 
Now the velocity of the dog is to that of the hare as 3 to2. How 
far does each run? 

12°. The same conditions as in the preceding only that the hare 
runs obliquely from the hound at half a right angle. 

Ans, The hound runs 60 (,/2 + 4/205), 
the hare 40 (,/2-+ ./2°05), rods. 

13°. Given the perimeter 4a, of a rectangle (right angled par- 
allelogram) and the diagonal, d, to determine the figure. 

14° Given the perimeter, na, of a right angled triangle, and the 
perpendicular, a, let fall from the right angle on the hypothenuse, 
to determine the triangle. 


Let the segments of the Reraiiennae be re- 7 3- Ne 
presented by x and y; then shall we have wee Be, 
ef vs 

r+y+ (2? +a)" +(y?+ @)* =na, Fig. 29. 
and = a * which ratio may be put = z; 
\ a 
then z will = az and y= >? 
1 
2 


2 a 
whence az +5 + (az? + a’) + ( S + a’) i na, 


or zt— * 4 (2 41P4 <(2+t=a; 


EXERCISES. 87 
$ 
ee ie ie Je +1) il ih etal 


and eA Y Dg Vp opin ced ie Goal oP atest ae i ea 
ee z 2 


or (n+ 2) 2-4 Fa 
“iene n’ ez=-—f], 
m+ 2 
n n‘ n' 
ss. (58),, 2* On 2 °Z+ (4n 4) au ap 16(n +1)? —1 
_ m—16 (n+ 17 
on, 10nd) 
a 
Balt Ie ey 
aft Cente (n+l) 
: 4(n+1) Tat) 
se ee (Ce ae 
‘ = 4(n+1) 
and y 2G or eet apa sO ot ayienis, 
a(n +1) 


15°. Given the two lines, m, n, drawn from the acute angles of 
a right angled triangle to the middle points of the opposite sides, 
to determine the triangle, 


16°. In a triangle, given the segments a, b, of the base formed 


by the perpendicular let fall from the vertex, and the ratio, m : n, 
of the oblique sides, to king the triangle. 


(a+ b) (a—b) (a + b) nay 
rH | eae: (m—n) ee" of 
17°. In a triangle, given the base, b, the eRe p, and 
the difference, 2m, of the oblique sides, z + m, x — m, to determine 
these sides. Ans, r+m=? zr—m=? 


18°. In a triangle, given the segments, a, b, of the base made 


, ees Re sides. 


88 EXERCISES, 


by the line bisecting the vertical angle, and the sum, 2m, of the 
obligue sides, to determine the triangle. 


Zbm 


19°. The same as in the preceding, ie the difference of the 
oblique sides is given instead of their sum. 

20°. Given the three sides of a triangle, 12, 18, and 20 chains, to 
find the perpendicular let fall on the last-named side from the op- 
posite angle. Ans. 1066 chains, 


21°. Given the base, b, and altitude, A, of a triangle, to determine 
the inscribed rectangle, the base of which shall be to the altitude as 


m to n. Whe mbh nbh 
Ans. The sides of the rectangle are Rea an 


22°. The same as the preceding, except that the sum of the sides 
containing the rectangle is given, 


23°. The same, only the difference is given. 


24°, Being on the bank of a river and wishing to ascertain its 
breadth, for the want of instruments I set up a stick at A and then 
take 25 measures with a rod happening to be at hand, from A ina 
straight line to B, where I set up a second stick; I then take 5 
measures of the same rod directly back from A in line with X, an 
object on the opposite bank, to C; also from B backward in line 
with X, 5 measures to D; and finally I find BC to be = 25 mea- 
sures and a remainder, also AD =26-+-a remainder. Now to de- 
termine these remainders, I proceed as follows : 
For the fractional part of BC, I apply to the mea- —; ae Ks! 
suring rod, m, a second rod, 7, = the excess of BC 
over 25m, and find r contained in m once, with a 


Fad 


Fig. 30. 
remainder, 7 ; therefore — are | ree Again, applying 7’ to r, I 
find it contained once with a remainder, 7’; .°, 4 =] Le . Ap- 
plying r” to vr’, I find it contained 4 times with a esate P's 
Radic wat 


, 7’ . 


} 


Finally, I find r” contained in r” just twice ; 


EXERCISES, 89 


Whence ee ee 
"14> 14— 14— 
{ 4 ees 
rs af 
Za 1 a 1 7 1 
1 oe 1+ _ 1+ = 
seis i Pe, er 1 
a on * jamie 
7" 
Be ge 1 u 1 Ve ee 
m i. F sies Olde: 
m 1+ 1 i¥ 1 par 
1 2 11 
1+—; rt 
ple Bory 
*. BC =25'55; 
by a like process I find AD = 26+ a 26‘65. 
1+ i 
1+ i 
I+— 


We have then (135) putting AX =z, BX = y, 
25'55? — 25? -y? — 25? 
eininatate S mate li Bes 


t@52 __ 2 2 re 
rn ae 25 nite - md aint 


i. r= y= : 
25°, Given the hypothenuse of a right angled triangle = 35 rods, 
and the side of the inscribed square = 12 rods, to determine the 
triangle. Ans. base = 28, or = 21; 
perpendicular = 21, or = 28. 


SECTION FOURTH. 


Comparison of Plane Figures: 
PROPOSITION I. 


Rectangles are to each other as the products of their di- (143) 
MeENSIONS. 


We will take the general case at once, and sup- 
pose the containing sides, a, a’, of the rectangle A 
are incommensurable with 0}, b’,those of B. Let 
b be increased by zx, and 0’ by z’, soas to make 
6+2z commensurable with a, and b'+ 2’ with a’. 
Suppose, for instance. that while a is divided into 
m parts, that b-+-z contains n of the same parts, 
or that (2,) 


SU, 

ben” 
and that, while a’ is divided into m’ equal parts, which may be dif- 
ferent from those of a, b'+ x’ contains »’ of the same parts, or that 

a ™ 
Digest 0.7. 
Through the points of division draw lines parallel to the sides of 
the rectangles and denote the part added to Bby X. Then will 
the partial rectangles be all equal (95) ; and the number constructed 
on the base a being = to m, and the number of tiers corresponding 
to the divisions in a, = n’, we have mm’ for the total number of 
partial rectangles in A; so the rectangle B+ X contains nn’ of the 
same. Hence (2,) 
A: B+ X=mm': nn; 

but mm':nn'::aa’:(b+2)(b' +2’), 


since a, a’; b+ 2, b'+ 2’, are as their measures m, m'; n, n’'; 


here iB teen Pa) (P +2") _ 66 + br tb arte 6 
A aa aa 
BX bh be’ +24 e2' | 
He Demo. ag 
BB bb 
. (63) be Q. E. D 


RECTANGLE AND TRAPEZOID, 91 


For it is obvious that X depends for its value upon z, 2’, decreas- 
ing as these decrease so as to become nothing if these were to be- 
come = 0, and that z, 2’, being less than the parts into which a, a’ 
are divided, may be made less than any assignable quantity, by suf- 
ficiently increasing the points of division. Therefore 
xX a+ bz’ ional ex’ 
A aa 


r! 


may be made less than any assignable quantity. 

Cor. 1. The rectangle is measured by the product of its (144) 
dimensions, For suppose A to become a measuring square, 
or that while A = 1 is taken as the unit of surface, a= a’ = 1 | i 


; I 
becomes the linear unit, we have Fig. 31s. 


Cor. 2. The right angled triangle is measured by half (145) 


the product of its base into its altitude. Thus T = 40h. “ 


Fig. 313. 


PROPOSITION II. 


The Trapezoid is measured by half the sum of its par- (146) 
allel sides multiplied into their perpendicular distance. 


Let a, 6, be the parallel sides and from their ex- b y 
tremities drop the perpendiculars h, h, on a, b, pro- ; “sh i 
duced, forming the rectangle Nines 

X4Trp+Y, : : 
aaa Fig. 32. 


composed of the trapezoid Trp and the right angled triangles X, 
Y, having the bases 2, y. We have (144), (145), 


X+ Trp+ Y= (r#+a)+h=4+2(c4+a) eh 
=7(t+a+b+y) eh, 
and X + Y=4trh+4yh; 
Trp=t(a+b)eh. Q.E.D. 
Cor. 1. The parallelogram is measured by the product (147) 


of its base into its altitude. For, when b=a, the , « 
trapezoid becomes a trapezium or parallelogram, BpeNOESP 
and Trp =+t (a+b) eh, Fig. 320. 


becomes, P=4t(a+a)+h=ah. 


92 PARALLELOGRAM AND TRIANGLE, 


Cor. 2. The triangle is measured by half the predees of (148) 
its base and altitude. 
For Trp =4(a+b)-h, hi 


becomes T = tah. hid 
Fig. 323. 


Cor. 3. Parallelograms are to each other, and triangles (149) 
are to each other, as the product of their bases and altitudes. 


Cor. 4. Parallelograms of the same or equal altitudes are (150) 
to each other as their bases. 


Cor. 5. Triangles of the same or equal altitudes are to .(151) 


each other as their bases, f 


Fig. 324. 
Cor. 6. Parallelograms of the same or equal bases are to (152) 
each other as their altitudes. 


Cor. 7. Triangles of equal bases are to each other as (153) 
their altitudes. 


Cor. 8. Parallelograms of equal bases and altitudes, are (154) 


equivalent, WT 


Fig. 325. 
Cor. 9. Triangles of equal bases and altitudes, are equiv- (155) 


alent, Ww 


Fig. 326. 

Cor. 10. Parallelograms, or triangles, between the same (156) 
parallels, and on the same or equal bases, are equivalent. 

Cor. 11. Conversely, if parallelograms or triangles, stand- (157) 
ing on equal bases in the same straight line, and having their ver- 
tices turned in the same direction, be equivalent, the line of ver- 
tices will be parallel to the line of bases, 

For if the triangles, having the bases a, a, and V- Vv" 
the perpendiculars p, p’, be equivalent, there results |, /\ JN w 

sap = tap, me ee 
ig. 327 
P=P; 
whence the line of vertices VV’ is parallel to the 1 of bases BB’. 
Cor. 12. The parallelogram is double the triangle of the (158) 


same base and altitude. 
[Compare (147) with (148).] Vs 
a@ 


Fig. 328. 


PARALLELOGRAM AND TRIANGLE. 93 


PROPOSITION III. 


Two triangles, having an angle of the one equal to an (159) 
angle of the other, are to each other as the products of the sides 
about the equal angles. 


Let the equal angles of the triangles A, B, be 
made vertical, and join the extremities of the sides 
a, 6, forming the triangle X; then (151) 


Ul 


wo 


I 


S|/& ofa ole 


and 


| 


~ 


Q. E. D. 


~ 
e 


byl bm byl be be 


Cor. 1. Equiangular parallelograms are to each other (159,) 
as the products of their dimensions (158). 
Cor. 2, Similar triangles are to each other as the squares (160) 


of their homologous sides. For, then we have ZI 
(122) 
a’ a. a <a 
ae ak 3 
A> Cee a a Oe OE May dias 
Robbe Dele | ioe eet 


Cor. 3. If figures, resolvable into the same number of (161) 
similar triangles, be constructed upon the three sides 
of aright angled triangle, that on the hypothenuse 
will be equal to the sum of those described upon the 
other two sides. For we have 
Davey CoP BEC EE a 
Fie or and ay © oes: RE wer re =4 ae ai ee eer 

Scholium. It is obvious that (132) is but a particular case of 
this more general theorem. 


EXERCISES. 


1°, Demonstrate (132) by turning two squares into one. . 
SLY 
Fig. 34. 

2°. On the oblique sides of any triangle describe parallelo- 


94 EXERCISES, 


grams, any whatever; andon the base, construct a par- 
allelogram having its sides parallel to the line drawn tL» 
through the vertex and the intersection formed by pro- 
ducing the sides of the parallelograms described on the 
oblique sides of the triangle, and terminating in these same sides. 

Prove that the parallelogram on the base of the triangle will be 
equal to the sum of the other two; and that, if the vertical angle 
become right, and the parallelograms squares, (132) will be 
proved. 

3°. It is required to straighten the line ABC, sepa- 
rating the estates of two gentlemen, by aid of the af 3 
cross alone. 

Draw BD parallel to CA, then DC will be the re- 
quired line. (Why 2) 


4°. Straighten the line in figure 36,. MA 


5°. It is required to change the direction of the line 
AB, so that the extremity A shall be at A’, still retaining r 
Fig. 363. 


Fig. 35. 


Fig. “36. 


the same amount of land on each side, 


6°. To turn a quadrilateral into a triangle, af bem 
either by the cross in the field, or by the right- peas 


angle straightedge on paper. Fig. 364. 
7°. To turn a pentagon into a triangle. Mears 
Fig. 365 
8°, To turn a triangle into a rectangle, JANN 
Fig. 366 


9°. To turn a rectangle, A, into a square, B. 


10°, Draw any polygon on paper, and find a square that shall be 
equal to it in area, 


11°. There is a well at P, in the side of a triangular Pp 
field; it is required to draw a line from P so as to Fat 


divide the field into two equal parts. Fig. 368. 
12°. What is the area of a parallelogram, 20 chains in length 
and 15 in breadth ? Ans, 30 acres. 


13°, The longer side of a rectangle is r times the shorter, and 


EXERCISES, — 95 


the area is = m*, or to a square having m for its side, What are 
its dimensions ? 
We have TreL= wv, 

f _ m _me/r 

< as SPRY aie ec 
and rz=m./T. 

14°. A surveyor would lay out a rectangular field of 20 acres, 
such that the length may be three times the breadth. Required, 
the dimensions, [Solve and prove. ] 

15°, The area of a parallelogram being represented by m?, and 
the base by b, how shall we find the height,  ?—given A, how shall 
we find 6? m? 


2 
Ans. h= = 7° How expressed in words ? 


16°. A man has 10 acres to put in a’parallelogram, having one 
side 12 chains. How wide must it be? Ans. 8 chs. 334 links. 

17°. If the area of a triangle be m?, and its base b, how shall 
we find its height, h? 


2 
Ans. h= — . [Enounce in words, ] 


18°. A surveyor would lay out a triangle of one hundred acres 
on a base of 25 chains. Required, the altitude. 
19°, A joiner has a board 10 feet long, 2 feet. wide at one end, 
and 2 feet 6 inches at the other. How many square feet in the 
board? (146) : Ans. 224 feet. 
20°. Given the sides of a triangle, 15, 16,17 chs, Required, 
the area. 
Taking 16 for base, we have (140) 
16: 17+15::17—15 : 2z, 
or Sx se cil s Sp. ods . 
“., (141) p = [(17+ 8+ 2) (17 —-8—2)]* =[9 f8 S27) 8 Y/OT 
Area = 8p = 24 VJ 21 sq. chs. = 11 — 54, acres, nearly. 
21°, What is the area of a triangle, the sides of which are 13, 
19, 34 chs, 2 
What difficulty arises in attempting to solve this problem? and 
why ? 
22°. Wishing to ascertain the area of the quadrilateral field 
ABCD ( fig. 6,), I avail myself of the measures in the third exer- 
cise of the preceding section, and find it to be nine acres and one- 
fifth, 
23°. In order to obtain the area of a pentagon ABCDE, I mea- 


96 EXERCISES. 


sure the sides BC, CD, DE, and find them BC =6, CD =4, DE 
= 10 chains; and, by aid of the cross, determine the perpendicu- 
lars AP = 3, AP, =8, AP, = 1158, let fall upon CB, DC, DE, pro- 
duced, The area will be found to be eight acres and two-fifths. 

24°, Given the sides of a triangle ABC, 

AB =30, AC = 40, BC = 50 chs., 

to cut off a triangle ABD (D being a point in AC) equal to 45 
acres. 

We find the area of the triangle ABC = 60 acres ;_ therefore (151) 

AD = 30 chs., the line required. 


25°. The triangle being the same as in the above, it is required 
to cut off the 45 acres by a line B’C’ parallel to BC—B' being a 
point in AB, and C’ a point in AC, 
We find (160) 
AB = 25:98, AC’ = 34°64 chs. 


26°. The same conditions as in the preceding, except that 30 
acres are to be cut off by a line PQ, P being a point in AB, 20 
chs. distant from A. Required AQ, measured off from A towards C. 

Join PC, then 

30 chs. : 20 chs. : : 60 acres : APC, .. = 40 acres; 
and 40 acres : 30 acres :: 40 chs. : AQ, .. =30 chs. 

27°. To draw a line through a given point in the plane of a given 
angle so as to include a given area. 

Let the given point P be embraced by the sides 
of the given angle CAB. Suppose the problem 
solved, and that YPX is the required line: we are 
to find z= AX, a portion of AB. As the position 4@—=x 
of the point P, in regard to the lines AB, AC, is Pigiet 
given, we may suppose the perpendicular p, let fall from P upon 7 
and a, the distance of P from the line AC, measured parallel to 
AB, to be known. Further, let the area of the triangle AXY be 
denoted by m’, that is, by a square whose side is m. 

We have, y being a perpendicular let fall from Y upon a, 

ry + p) = 2m’, 


and 


x 
a 
fe L—a 
or ps = 
¥ 


EXERCISES, 97 


ee eee 


pi hag uny tae Ty 


oy +p)=2( ee +p) = (BS + px = 2m?, 


wo == @& t—G 


or par + px? — par = 2m*r — 2m*a, 


2m? 2m*a 
zr? ezrt=-— 9 
P P 
a m*\?> m* 2m?a. m 
r?—Be—eer (~) = — — —_ = — (m*? — 2pa 
p) pp pi pa) 
2 
o— T= = (m* — &pa) ; 
2 1 
and =¢= - +7 (m* —2pa)* = 5 [m + (m? —2pa)?}. 


4 
Since (m*? — 2pa)* is < m, the two values of z, 


z= — [m+ (m* — 2pa)*}, 


and z= = [m — (mn? — 2pa)*], 

are both plus; and, therefore, equally applicable to the problem, 

as shown in fig. 37,, where we have : 
t= AX, zr = AX’. P 


We observe here since 
the tr. AXY = m?= AX’Y’, 
that subtracting the quadrilateral AX’PY from both, there results 
the triangle XPX’= YPY’. 
But if we take the difference of the values of X, we have 


XX says 7 + 2 (m? — 2pa)*, 


Fig. 372. 


which is the solution of the following problem : 

27°,. Through a given point, P, in the side of a given triangle, 
AXY, to draw a line X’Y’; terminating in AX, AY, produced, if 
necessary, so that the areas PXX’, PYY’, shall be equal. 

We remark that m? must be at least as great as 2pa; for, were 


m?* < 2pa, (m? pa)" would be imaginary (6;), and the values of 
xz, © = AX, x = AX’ unreal, or the problem impossible. That is, 
the area to be cut off cannot be greater than the double inscribed 
rectangle standing upon a. 


98 EXERCISES, 


We have here an example of a quantity capable of a minimum, 
or least value ; and, in order to find this minimum, it is obvious 
that we have only to solve the quadratic and put the part under the 
radical equal to zero, 

Thus m? — 2pa = 0, 
gives m? = 2pa, for the minimum of m?, 

If we inquire what is the greatest value of m? in this problem, 
we shall find it infinite, or that m? has no maximum. 


ax 
[See the value of z and fig. 37;.] ee 
x —> 
If now we subtract from a by insensible degrees, Fig. 373. 


or, as it is commonly expressed, diminish a according to the Law 
oF ConTINUvITY, a will at length become = 0, 


m mS 
when te 5 [m + (m? — 2pa)*], 


2 
gives BD ee ee mdr oe 

Fig. 374. 
where the first value only is applicable. 


Continuing the same motion, a will evidently become minus, and 
the point P will at the same time take up its position on the left of 


the line AC, or without the angle, and we shall have . 
ry <3 A 
z= = [mt (m? + 2pa)*] : rs 


Fig. 375. 
where both values of x are applicable, only that the second, being 
minus, requires the area m? to be laid off on the left. 

If, in like manner, we make p pass through the value 0 it will 
change its sign, and P will take up a position be- 
low the line AB, 


os 1 
whence Rae ss [m + (m?— 2pa)*], 
of 1 
becomes Bae et ee 
m 1 
or t=—~ [— m = (m* + 2pa)?], 


where both values are applicable, as indicated in the figure. 
If, at the same time, we make a and p both minus, there will 
result 


B 
r= — [m+ (m?—2+—p+—a)4), 


EXERCISES, 99 


or r=— $ [m = (m? — Qa)? ], 


where we have the same values as in the original solution, only 
changing + 2 into —2z, as we evidently ought to do, since z is 
measured from A in an opposite direction. This is obvious also 
from the comparison of the figures 37, and 37, where all the parts 
of the one correspond to all the parts of the other. The figure 37, 
will have the same double construction that 37 has in 37,. 

What would be the result of making p=0? what, when p= 0, 
a=0? . 

The problem we have just been discussing is admirably adapted 
to show the correlation of algebraical signs and geometrical figures. 
This problem will also enable us to divide any polygon into any 
required parts. 

28°. The parallel sides of a trapezoid are a, 6, and the altitude, 
h, it is required to cut off an area m? adjacent to b by a line paral- 
lel thereto. What will be its breadth? 


bh 2hm? bh \?|% 
pine e=— + [57+ (3) J 
29°. Given the sides of a triangle, ABC, viz., AB=10 chains, 
AC =8 chains 43 links, BC = 470; also the position of a point, 
P, viz., distant from AB by 1‘80, and from AC by one chain, Re- 
quired to draw a line through the point P, that shall divide the tri- 
angle into two equal parts. : 
The student will solve and verify. 
30°, What is the greatest rectangle with a given perimeter ? 
We may denote the sides by a+ x and a —z, then will the peri- 
meter be = 4a, and we shall find 
z= (a? — m?)?, 
m* denoting the area; whence it is obvious that m? cannot exceed 
a*; and therefore that z= 0, when the area m? is a maximum, or 
that among rectangles of the same perimeter, the square is the 
maximum. 
31°. What is the maximum triangle that can be constructed ona 
given base, and of a given perimeter ? 
Ans. The triangle must be isosceles, 
32°. Given the area and diagonal of a rectangle, to determine its 
sides, 


33°. Let ABCD be a quadrilateral field, of which the sides AB, 


q 


100 EXERCISES, 


BC, CD, DA, are, respectively, 16, 34, 30, 29 rods in length ; also, 
let the diagonal BD = 374 rods. Itis required to divide the field 
into two equal parts, by a line cutting the opposite sides, AB, CD, 
so that the ratio of the segments of the one shall be equal to that 
of the corresponding segments of the other. 
Ans, AX=? DY =? 

[Produce AB, CD, until they meet, and consult (140), (141), 

(132), (159). ] 


BOOK THIRD. 


PLANE GEOMETRY DEPENDING ON THE CIRCLE, 
ELLIPSE, HYPERBOLA, AND PARABOLA. 


SECTION FIRST. 


The Circle. 


Definition 1. The circle is a plane figure described by the revo- 
lution of a straight line of invariable length about one of its extrem- 
ities as a fixed point. 

Def. 2. The describing line is called the Radius [rod] of the 
circle, the fixed point the Centre, and the curve line that bounds it 
its Circumference. : 

Cor, All radii, or lines drawn from the centre to the cir- (162) 
cumference, are equal to each other. 


PROPOSITION I. 


Angles at the centre of the same or equal circles, are to (163) 
each other as their subtending arcs ; and the corresponding sec- 
tors have a like ratio. 


In the first place, let the angles be commensurable. For exam- 
ple, suppose the angle AOB to contain the angle 
BOC twice ; then it is manifest that in applying the 
angle BOC twice to the angle AOB, the point C 
will fall on C’, the middle point of the are AB, since x 
(162) OC’=OC. Therefore the are AB contains 
the arc BC twice; and, in the same way, the sector 
AOB is double the sector BOC. 

“. ZAOB: ZBOC:: are AB: arc BC: : sec. AOB: sec. BOC, 
each being as 2 to 1. | 


Fig. 38. 


102 SECTORS. 


So in general, if a, 6, be any commensurable an- 
gles, or such that when a is divided into any num- 
ber, m, of equal parts, b shall also be exactly divisible 
into some number, n, of the same equal parts ; then 
it will follow, by superposing one of these equal Fig. 389. 
angles m times upon a and nv times upon 3, that the 
corresponding arcs A and B will be divided into m and n equal 
arcs. The same of the sectors K, L. 

a:b6:: A: B:: K: L, being asm to n. 


Finally, let us take the most general case, or that 
of incommensurability, where a and b having no 
common measure, are incapable of being divided into 
the same equal parts. Let. 0d be increased by the 
angle z, so that b-++ x shall be commensurable with 
a, then will the corresponding are B+ X be com- 
mensurable with Aj; and, from what has just been proved, there 
results the proportion 


& p 
Fig. 383. 


b+2 Bx 
i bitten Ae ou! 
Ie re eon «See 
bs naa nrriats taste 
bus B iv A 
, (63) aaa TB Bere oti A: B, 


and the same is equally true of the sectors. Q. E. D. 
Cor. 1. In the same or equal circles, arcs may be taken (164) 
as the measures of their angles at the centre. 
Thus, if 6 be taken for the unit of angles and B for 
that of arcs, 


the proportion Se al : Fig. 384. 
b «cB 

becomes ban = 
1 1 

or , Gite JA § 


Scholium. As the right angle seems the most suitable for com- 
paring angles, so its measure, the Quadrant, or quarter circumfer- 
ence, would appear to be the appropriate unit of arcs, and for this 
purpose the French have sometimes employed it, dividing the quad- 
rant into a hundred equal parts, which they called degrees: but 
custom has established a different unit, the jth part of the quad- 


SECTORS, 108 


rant, which is denominated a degree, and written 1°. The degree 
is divided into 60 minutes, marked 1’ and the minute into 60 sec- 
onds, written 1”, 2”, 3”, .... 

Cor. 2. A quarter circumference is the measure of aright (165) 
angle, a semicircumference of two right angles, and a circumfer- 
ence of four right angles. 

How many degrees in half a quadrant? in one third of a quad- 
rant? in4d? 427424247247 4? 

Cor. 3. In the same or equal circles, the greater arc (166) 
subtends the greater angle at the centre, and, conversely, the greater 
angle is subtended by the greater arc. 

Cor. 4. In the same or equal circles, equal arcs sub- (167) 
tend equal angles at the centre, and the converse. 

Cor. 5. In the same or equal circles, the greater arc sub- (168) 
tends the greater chord, and conversely, the greater chord is sub- 
tended by the greater arc. 


For, let the arc B> A, .. (166) 7 b>a; «. 
(116) chord d>c. Conversely, let chordd>c, 7 


* Zb>a;., arciBiu. ; 
é 


Fig. 385. 
Cor. 6. In the same or equal circles, equa] arcs subtend (169) 
equal chords, and the converse. 
Cor. 7. The Diameter, or chord passing through the (170) 
centre, is the greatest straight line that can be 
drawn in a circle. Why? 


Fig. 386. 
Cor. 8. The diameter bisects the circle, and its circum- (171) 


ference. 


PROPOSITION II. 


An angle inscribed in a circle, is measured by half tts (172) 
subtending are, 


Let AOB be any diameter, and C any point of the circumfer- 
ence: join CA, CO, then (94), (162), (101), 


104 INSCRIBED ANGLES. 


Z COB = OAC + OCA =2 # OAC; 
.. (164), measure of 7 OAC=meas, of + 7 COB 


=4arc CB; 
so meas, of 7 DAB=#4 arc DB, 
and meas, of 7 EAB=4 arc EB; 
.., by addition and substraction, : 
meas, of / CAD = 4 are CD, [Centre where ?] Fig. 39. 
and meas. of 7 DAE =¢t are DE. [2] Q.E. D. 


Cor. 1. In the same or equal circles, angles at the cir- (178) 


cumference, subtended by the same or equal arcs, 
are equal. ne " 
; Fig. 392. 
Cor, 2, Of angles inscribed in the same or equal circles: (174) 
1°. An angle subtended by an arc less than a semi- (fig. 39,.) 
circumference is less than a right angle ; 


2°. An angle subtended by an arc greater than a (fig. 39,.) 
semicircumference is greater than a right angle ; . 


3°. An angle subtended by a semicircumference 
is a right angle. 


Fig. 393. 
Cor. 3. The sum of the opposite angles of a quadrilate- (175) 
ral inscribed in a circle is equal to two right angles. (fig. 39.) 


eee igh. 

Cor. 4. A Secant line is always oblique to the diameter (176) 
drawn through either point, in which it cuts the circumference ; 
and, conversely, a line drawn through the extremity of a diameter, 
and oblique to it, is a secant line, or cuts the circle. 

For, let SS’ be any secant line, cutting the cir- “ 
cumference in any points A, X, and draw the diame- 
ter AB; then the 7 BAX is < |, being measured 
by the are BX < semicircumference. Conversely, 
let the line SAS’ be oblique to the diameter AB, 
and suppose that BAS’ is the acute angle, then will 
SS’ cut the circle.. For, take the point X in the . Fig. 394. 


SECANTS AND TANGENTS, 105 


semicircumference on the same side of the diameter AB with the 
angle BAS’, so that the angle BAX shall be equal to the angle 
BAS’, which may always be done, since, by taking the point X 
nearer and nearer to A, the angle BAX may be made any what- 
ever less than a right angle; then will the line AX coincide with 
the line AS’, and the point X will be a point of the line AS’, 
Therefore AS’ will cut the circumference a second time in X, and 
be secant to the circle. 


Cor. 5. A Tangent, that isa line which touches a circle (177) 
without cutting it, is at right angles to the diameter drawn through 
the point of tangency, and the converse. 

Let TAT’ be tangent to the circle at A; 
then will TT’ be. perpendicular to the diameter 
AB; for, if TAT’ were oblique to AB, it 
would be a secant line (176). Again, let TAT’ 
be perpendicular to AB, then is TT’ tangent 
to the circle; for, if AT were a secant, then 
would 7 BAT < +, which is contrary to the 
hypothesis, 


Illustration. Imagine a line to revolve about a 
point in the circumference of a circle, there will 
be but one position in which it will be a tangent, or 
in which it will not cut the circle. 


Fig. 39«. 
Cor. 6. The angle formed by a tangent and secant is (178) 
measured by half the included are. 
For the Z BAT,= +, is measured by 4 arc. BXA, (fig. 39;.) 
and Z BAX is eta by 4 arc BX; 
‘ [—]  Z XAT is measured by 4 arc XA. 


PROPOSITION III. 


The angle formed by the intersection of two secant (179), 
lines is measured by the half sum or the half difference of the 
included arcs, according as the point of intersection is within or 
without the circle. 


First, let the lines AY, BZ, intersect in X, a point within the cir- 


106 CORRELATION OF FIGURES, 


cleABYZY’; join BY, then the measure of Z BXA =the meas, 
of (BYX-+ YBX ) 
= the meas. of BY X-+ meas. of YBX, 
=4 arc AB+¢ are ZY, 
=+ (are AB-+ are ZY). 

Now let the point X glide along the 
line BX ZX’ until it take up a position 
X’ without the circle ; as a consequence Fig. 40. 
of this motion of X, the arc ZY will diminish, vanish, and finally re- 
appear measured in the opposite direction from Z, as ZY’. But 
this diminution of the arc ZY may be regarded as produced by 
subtracting from ZY, a quantity greater than itself; therefore, the 
remainder ZY’ must be a minus quantity, ZY being plus: 


Thus ZY +BY VY, w-ZY' = VY ZY, 
or — ZY'=ZY—YY'. 
Whence the measure of 7 BX’A = 4 (arc AB —arce ZY’). 
Q. E. D. 


Remark. The second part of the demonstration may be made 
independently by joining AZ; for the measure of 7 BX’A = meas. 
of (BZA — X'AZ) = 4 (AB — ZY’); but it is important to arrive at 
a principle by the aid of which we may be enabled, as above, to 
comprehend the mutual dependencies of all the particular cases of 
a general proposition. 


PRINCIPLE.—Whenever a Geometrical Magnitude and its Algebrai- (180) 
cal Representation can be made, by CONTINUOUS Diminution, to pass through 
the value nothing, and, by the same “Law of Continuity,” to re-appear—then 
will the Geometrical Magnitude be opposite in position, and its Algebraical Re- 
presentation be affected by the contrary sign. 


For it is evident that the magnitude, whether line, surface, solid 
or angle, on passing through zero, will change direction, and it is 
equally obvious, from the theory of algebra, that any algebraical 
expression, as ad — b, by which it may be represented, will at the 
same time change from plus to minus or vice versd. 

Scholium. Hence magnitudes, as two lines measured in oppo- 
site directions, are usually affected by contrary signs, but not al- 
ways. Thus, two radii of the same circle, though measured from 
the same point, the centre, in opposite directions, constituting a 
diameter, are not to be regarded the one as + and the other as — ; 
since the one cannot be derived from the other by passing through 0. 


INTERSECTING CHORDS. 107 


Cor. Parallel lines intercept equal arcs; and, conversely, (181) 
lines intercepting equal arcs are parallel. 

For the lines AY’ and’ BZ being parallel, the angle (fig. 40.) 
X'=0, .*. meas. of X’=0, or 4 (arc AB—are 


2 B 
“ZY )=0, « are AB= ZY. . 
Again, if AB = ZY' then 7 X' = 43 (are AB — are x r, 
ZY’) =0, and .. AY’ is parallel to BZ. This 


corollary might have been placed under the i Fig. 402. 
ceding proposition. How? Join AZ. 


PROPOSITION IV. 


When two intersecting lines cut a circle, the product of (182) 
the segments of the one, included between the point of intersec- 
tion and the circumference, is equal to the product of the cor- 
responding segments of the other. 


Let the chords AY, BZ, intersect in X; join BY, AZ, (fig. 40.) 
then the triangles AXZ, BXY, are equiangular and similar, whence 


the proportion “ 
AX BX 
XZ XY? 5 AX @ XY = BX e XZ, 


and the theorem is proved when the point X of intersection falls 
within the circle. Now let the point X glide along the line 
BXZX’ till it takes up a position X’ without the circle; then will 
the segments XY, XZ, decrease, pass the value nothing, and re- 
appear as X'Y’, X'Z, measured in the contrary direction ; 
(180), AX « XY = BX « XZ, 

becomes AX’. — X'Y’= BX’. — XZ, 
or AX’. X'Y'=BX’.X’Z. Q.E. D. 

Cor. 1. Through three points not in the same straight (183) 
line, a circumference of a circle may be made to pass, and but one, 

Let A, B, Y, be three points not in the same straight (fig. 40.) 
line; join BY, YA, then draw BX’ at pleasure, and make the 
angle YAZ = YBZ;; it follows, from what has been proved, that 
Z will be a point of the circumference. In the same way any 
other point may be found; therefore the position of the three 
points A, B, Y, determines the positions of all the points of the 
circumference. 

Cor. 2. A straight line cannot cut the circumference of a (184) 


108 CIRCLE AND TRIANGLE INSCRIBED. 


circle in more points than two; since through three points of a 
straight line no circumference can be made to pass. 

Cor. 3. If from any point without a circle a tangent and (185) 
secant line be drawn, the portion of the tangent intercepted between 
that point and the point of tangency will be a mean proportional 
between the segments of the secant line. 

For, let the points A,Y’ approach each other so as to (fig. 40.) 
become united in ,T,,,; then will the chord AY’ 


vanish, and X’Y’A become a tangent X’ ,Ty.; <4 x! 

further X'Y’ will = X'A=X’ ,T,.; ee 
“. XB +s X'Z= XA. XY’ =X', Ty, + X iT, de 

Corea D te Py, fs rk Fig. 403. 


Cor. 4. From the same point two equal tangents can be (186) 
drawn to a circle. | 

[Unite B, Z.] 

Cor. 5. If a line be drawn bisecting the vertical angle (187) 
(a, 6) of a triangle, and terminating in the base, the product of the 
oblique sides a, 6, will be equal to the product of the segments m, 
n, of the base, increased by the square of the bisecting line c. 

Suppose a circle described about the triangle, pro- S 
duce the bisecting line c, so as to form the chord LBS 
c+ x, and join the extremities of c+ xz and bd; then, 
by similar triangles, we have 

a” c+e 
ore otra 
ab=c?+cxr=mn-+ c?, 


Fig. 404. 


PROPOSITION V. 


The continued product'of the three sides of any triangle (188) 
ts equal to its double area multiplied into the diameter of the cir- 
cumscribing circle, 

Let ABC be any triangle; suppose a circle cir- 
cumscribed, and draw the diameter BD, also CP 
perpendicular to AB; then, by similar triangles, 

CP CB AC -CB 


AGA Bei Ss oApas 
AB- AC. CB 

A =aith e@ =: 

ZAABC=CP.- AB BD : 


or AB-+BC-CA=24 ABCXBD. Q.E.D. 


EQUATION OF THE CIRCLE. 109 


PROPOSITION VI. 


To find the Equation of the Circle referred to Rectangular 
Coérdinates. 

Let OX, OY, be two lines, intersecting each 
other at right angles in O; also let 0, situated at 
the distance 56 from OX and a from OY, be the 
centre of any circle; further, let P be any point of 
the circumference, distant from OX and OY by 
the variable lines y and z; then will the radius, 7, 
be the hypothenuse of a right angled triangle, of which the base 
will be z —a, and the perpendicular y — b, whence 

The Equation of the Circle 
is (y— 5b)? + (4@— a)? =r%, (189) 

Scholium. The variables zx, y, are called Codrdinates of the 
point, P, or of the circumference, when spoken of together ; when 
one is to be distinguished from the other, y is denominated the 
ordinate and zx the abscissa; OX, OY, are called the axes of co- 
ordinates, OX is the axis of z, OY that of y—O is their origin. 

If the centre o of the circle be on OX, the axis of zx, b = 0, and 
the equation (189) becomes 

y?+(x—a)?=r’. (190) 


to. 


‘ Fig. 422. 
If, in addition tob=0, we make a=Q, the centre, 0, will be 

transported to the origin, O, and the equation becomes 
y? + 7? = 72, (191) 


Fig. 42. 


Resolving (191) in reference to y, we find two 
equal values of y for every value of z, 


y =: (r? — 2°)8, 


i.e. y=+(r? — x)?, or y = — (r? — 2)?, iN 
Fig. 423. 
Now, if we diminish the arc AD, its ordinate AB =y diminishes 
also, and becomes=0 when AD=0; finally AD reappearing 
as A’D measured in the opposite direction, its ordinate A’B reap- 
pears, likewise measured in the opposite direction from B, and 


110 INTERSECTING CIRCLES. 


is therefore = — y, according to the principle laid down in (180). 
Whence, 

Cor. 1. The circle is perfect:y symmetrical, any diameter, (192) 
as DD’, bisecting allthe chords to which it is perpendicular[ fig, 42, ]. 

Cor. 2. The perpendicular which bisects a chord passes (193) 
through the centre of the circle and bisects the arc. 

Cor. 3. The greater chord is less distant from the centre, (194) 
since the semichord y increases as x diminishes; and the diameter 
is, therefore, the greatest straight line that can be drawn in acircle. 

By comparing equations (189), (190), (191), with the figures 
(42), (42,), (42;), which illustrate them, it is obvious that the co- 
ordinates y, x, represent, in general, totally different quantities in 
these different forms, so that one cannot be combined with another, 
as in ordinary algebraical operations. But if the circles intersect, 
as in figure 42,, then the point of intersection, I, being the same 
for both circumferences, the codrdinates of this point, y;, 2, will 
satisfy the equations of both circles, and from (191) and (190) de- 
noting the different radii by 7,72, we have 

¥ +2P=7’, 
y?+(%—a) =r; 
whence 2azr;,— a* = r* — 72. 
Resolving the last equation in reference to a, we find 


@=2,4(2°+ 7,7 — ry? 
whence, for any compatible values of z;, r., r, we find two values 
for the distance of the centres, a; or, for any given radii and a giv- 
en chord, the circles can be made to intersect in two ways. 
Ifr,<r, the values of a will be both+, the one greater than 
z;, the other less, and the centres will be 


on opposite sides of the ordinate, Yis 


or on the same side, accordingly. 


Fig. 425. 


INTERSECTING CIRCLES. 111 


aa plus, ] fas rep. in s 


If r, >, the values 
of a will be, 


cc other 
minus ; and 
Fig. 427. 
if : Qt 22) Fe 
we take — 7; instead of +2, @ = 27,4 (277+7.2—r?)?, 
4 + 
becomes a = — x, + (22+r,2—r?)*, or —a=2, (22 +7 —1?)? ; 


whence we have the same constructions over again, only on the op- 
posite side of the origin. 


2 t ps2. pe, 2 
If we substitute the value of x = oar in y?+27%=r?, 
aot 2 2\ 2 t 
a + pile (dibs ids ] 
we find y; . ( a 


whence it follows that the circumferences intersect in two points, 
I, I', equally distant from the middle, B, of their common chord ; 
and .*, that, as long as y, has a real value, the triangle OI, must be 
possible. From what has been demonstrated, we have the follow- 
ing corollaries: 

Cor. 4. Two circles of given radii will have for a com- (195) 
mon chord four positions of intersection. 


Fig. 42s. 
Cor. 5. The line joining the centres of intersecting cir- (196) 
cles, is perpendicular to, and bisects their common 
chords, 


Fig. 429. 

Cor. 6. In order that two circles intersect, of the three (197) 

quantities, their radii and the distance of their centres, any two 
must be greater than the third. 


Cor. 7. The distance of the centres of two tangent cir- (198) 


112 EXERCISES. 


cles, is equal to the sum or difference of their radii, 
according as they touch externally or internally. 


Fig. 4210. 
Cor. 8. The line joining the centres of tangent circles, (199) 
passes through the point of tangency, and the converse, [ fig. 42,..] 


EXERCISES. 


1°. What is the greatest triangle that can be inscribed in a semi- 
circle? Ans. An isosceles triangle, standing upon the diameter. 

2°. What is the greatest rectangle that can be inscribed in a given 
circle ? Ans, A square, 

3°, What is the maximum triangle that can be inscribed in a 
given circle and standing on its radius? 

Ans. A triangle right angled at the centre, 

4°, How many boards 15 inches wide and an inch thick can be 
cut from a log 20 inches in diameter ? Ans. 5 ./7, 

5°. What must be the diameter of a water wheel to fit an apron 
2a feet across and b feet deep? 

6°. How much must a plank be cut out to makea felly 1 ft. 
6 in. long to a wheel 6 ft. in diameter, the measures being taken on 
the inside ? 

7°. What is the radius of the largest circle that can be cut from 
a triangular plate of silver, measuring 23, 3, 3‘5 inches on its sides ? 

8°. Three brothers, residing at the several distances of 10, 11, 
12 chains from each other, are to dig a well which shall be equally 
distant from them all. What must be that distance ? 

9°, The diameters of the fore and hind wheels of a carriage are 
4 and 5 feet, and the distance of their centres 6 feet. At what 
point will a line joining these centres intersect the ground, sup- 
posed to be a plane? 

10°, There are two wheels situated in the same vertical plane, 
and their centres in the same vertical line ; the largest, the centre 
of which is 10 feet below the floor, is 8 feet in diameter, and the 
smaller, the centre of which is 6 feet above the floor, is one foot in 
diameter. Where must we cut through the floor for the passage 
of the strap that is to embrace the wheels ? 

11°. The same as in the above, except that the strap is to cross 
between the wheels. 


EXERCISES, 113 


12°, The same conditions as in 10°, except that the centre of the 
lower wheel is 4 feet from the plumb line dropped from the centre 
of the wheel above the floor. 

13°. It is required to construct three equal friction wheels to run 
tangent to each other and to an axle two inches in diameter. What 
must be their common radius, and what the radius of the circular 
bed cut for them in the centre of a wheel? 

14°, If the top-masts of two ships, having elevations of 90 and 
100 feet above the level of the sea, are seen from each other at the 
distance of 25‘7 miles, what is the diameter of the earth ? 

15°. How far can the Peak of Teneriffe be seen at sea? 

16°. How far will a water level fall away from a horizontal line, 
sighted at one end in a distance of one mile, the diameter of the 
earth being estimated at 7,960 miles ? 

17°.* If AC, one of the sides of an equilateral] triangle ABC, be 
produced to E, so that CE shall be equal to AC; andif EB be 
drawn and produced till it meets in D, a line drawn from A at right 
angles to AC; then DB will be equal to the radius of the circle 
described about the triangle. 

18°. If an angle B of any triangle ABC, be bisected by the 
straight line BD, which also cuts the side AC in D, and if from 
the centre A with the radius AD, a circle be described, cutting BD 
or BD produced in E; then BE : BD :: AD: CD. 

19°. Let ABC be a triangle right angled at B; from A draw AD 
parallel to BC, and meeting in D, a line drawn from B at right 
angles to AC; about the triangle ADC describe a circle, and let E 
be the point in which its circumference cuts the line AB or AB pro- 
duced; then AD, AB, BC, AE, are in continued proportion. 

20°. Let ABC be a circle, whose diameter is AB; and from D 
any point in AB produced, draw DC touching the cirele in C, and 
DEF any line cutting it in E and F; again, draw from C a perpen- 
dicular to AB, cutting EF in H; then, 

ED*: CD? :: EH: FH. 

21°. Let ABC bea circle, and from D, a point without it, let 
three straight lines be drawn in the following manner: DA touch- 
ing the circle in A, DBC cutting it in B and C, and DEF cutting it 
in E and F; bisect the chord BC in H, draw AH, and produce it 
till it meets the circumference in K; draw also KE and KF cut- 
ting BC in Gand L. The lines HG and HL are equal. 


* “ Prize Problems,” Yale College, 1840 
8 


SECTION SECOND. 
The Ellipse. 


Def. 1. An ellipse is a plane curve described by the intersection 
of two radii, varying in such manner as to preserve in sum the 
same constant quantity, while they revolve about two fixed points 
as centres, 


PROPOSITION I. 


To find the Equation of the Ellipse. 


Let P be any point of the curve, 

formed by the intersection of the vari- 
able radii a+ u, a — u, which are equal 
' in sum to a constant quantity, 2a, and 
which revolve about two fixed points, Fig. 43. 
F, F’, distant from each other by the line 2c. Take the middle 
point O between F, F’, for the origin of rectangular codrdinates, 
_ the line passing through F, F’, being the axis of X. There 
results, 


(a+ u)? =y? + (¢+2)? 
and (a—u)? =? + (c—2)*5 
[+h  @+ur=ytor+z%, 


and, [—], au=cz, 


c 
or t= te Ee 
a 


2 
a+ — (Pat w=yr +e? + 2, 
or a?a? + c*z? = a*®y*? + a?c? + az? ; 
ri a*y? + (a? —c*)xz? = a’(a’ — c*), (200) 
the oanaten of the ellipse. 

Now, it is obvious that, if the curve cuts the axis of 2, y for that 
point will be reduced to nothing ; therefore, if we make y = 0, and 
denote by z,_, what x becomes for this value of y, we have (200) 

a . 0? + (a? — Oe a (a? — c’) ‘ 


Ty) = =a; hence 


AXES, 115 


Cor. 1. The ellipse cuts the axis of abscissas at equal (201) 


distances on the right and left of the origin, which x°' x 
i é =a. en SAS 
distane a Fig. 430. 


When z =0, the curve cuts the axis of y, but this condition 
gives (200) 
ays _g + (a — c%) +» OF = aa? — ) 5 
y, p= (C—&)*; hence, 
Cor. 2. The ellipse cuts the axis of ordinates at equal (202) 
distances above and below the origin, which distance, de- jy 
noted by Yreso 


b, = (a? —c*)z. 


Def. 2. The line MON = 2a, is denominated 
the Major Avzis or the Conjugate Diameter, 
and passes through the Foci,* F, F’; the line 
POQ = 20, is the Minor Axis or the Trans- 
verse Diameter, being perpendicular to the for- 
mer, [6 can never > than a.] 

If in (200) we substitute 5? for (a? — cc”), the equation of the el- 
lipse becomes 


Q 
Fig. 434. 


a’y? + b’2* = a’b*, or == |y (203) 


Ye 
r a 
where the constants are the semimajor ae iid axes, 
Cor. 3. The ellipse is symmetrical in reference to both (204) 
axes ; since, 


For every value of z, whether + or — 
% ae , 2 
we have y = + (b? — ans 2?)2, two equal values Mh il 


Fig. 435. 


of y, one +, the other —; 
and, for every value of y, whether + y or — y, 


a? 1 
we have tr =+ (a — oe y’)*, two equal values 


of xz, one +, the other —; hence, Fig. 436. 
Cor. 4. The major axis bisects all chords parallel to the (205) 

minor, and the minor axis bisects all chords parallel to the major. 
Cor. 5. The origin bisects all chords drawn through it, (206) 


+ Foci, plural of focus, fire-place. 


116 DIAMETERS, 


and is, consequently, the Centre of the Ellipse ; 
therefore these chords are Diameters. sa) 


Fig. 437. 

Cor, 6. Diameters, equally inclined to the major axis, (207) 
are equal; and the converse. 

Cor. 7. Any ordinate of the ellipse is to the correspond- (208) 
ing ordinate of the circle, described on the major axis, as the semi- 
minor axis is to the semimajor. 

For let y, Y, be corresponding ordinates of the 
ellipse and circle; we have (203) 


b2 
y= (a — 2%) 


and (2) Y?=a*—~z’; 
a es ae ee ee Fig. 43. 
Cor. 8. The circle, described on the major axis, circum- (209) 
seribes the ellipse. Hence, 
Cor. 9. The angle embraced by chords, drawn from any (210) 
point of the ellipse to the extremities of the major axis, 
is obtuse. EN 
[How 2] Fig. 439. 
Cor. 10. Any abscissa of the ellipse is tothe correspond- (211) 
ing abscissa of the circle described on the minor axis, as_ the semi- 
major axis to the semiminor. 
For, we have 


2 
r= as —y*), 
and X? = b?—y?; 
: Tis um ee Fig. 4310. 


Cor. 11. The circle described on the minor axis is in- (212) 
scribed in the ellipse. 

Cor. 12. The angle embraced by chords drawn fromany (213) 
point of the ellipse to the extremities of the minor axis, is acute. 

Def. 3. The double ordinate drawn through the focus, is denom- 
inated the Parameter of the major axis, and sometimes the Latus 
Rectum. 

To find the parameter we have only to make x =c in (200) ; 
whence there results 


PARAMETER AND ECCENTRICITY. 117 


a*y? __. + (a? — c*)c? = a*(a? — c?), 
a*—c* .2b% 4b? (2b)? 
oo a 2a) hte 
or 2a : 26 :: 2b: Parameter ; i. e. 
Cor. 13. The Parameter is a third proportional to the (214) 
major and minor axes. 
Putting the parameter = p, and substituting in (203) we get 


Parameter = 2y,.,= 2 


y? = =e (a? — x2), (215) 


for the equation of the ellipse, in terms of the parameter and semi- 
major axis. 
We may transform (200) into 


a*y? = (a* — c*) (a? — x), 


2 
or y? = (1 = 5) (a* — 7”), 
or y* = (1 — e*) (a? — 2”), (216) 
putting e= < ‘ (217) 


Def. 4. We call e the Eccentricity because it expresses the 
ratio of the distance, c, of the focus from the centre to the semi- 
major axis, and thus determines the form of the ellipse, as round 
or flat. When the eccentricity is = 0, the ellipse becomes a circle. 
Equation (216) is that of the ellipse, referred to its eccentricity 
and semimajor axis, and is convenient in astronomy. 

It is sometimes desirable to have the equation of the ellipse, 
when the left hand extremity of the major axis 
is made the origin of abscissas. In order to 


this, we have only to substitute z—a instead of - 
zx in (203), as the new z exceeds the old by a, y 
remaining the same ; which done, there results, he ie 
2 
y= 2 (2ax — x”), (218) 


The origin might be transported to any other point, either in the 
curve or elsewhere, by changing the value of y as well as that 
of 2. 3 

Scholium. It is easy to show that any equation referred to rect- 
angular coérdinates, and of the form 


py + 92? = 7; 
is the equation of an ellipse; for we have 


118 TANGENT TO ELLIPSE. 


y? zr? a 1 
eo ery ee 
ae 
which will agree with (203), by putting 
b=, a? = oy 
Od q 
er r r 
buneniting b—./ (=) eceal a 


PROPOSITION II. 


A tangent to the Ellipse makes equal angles with the (219) 
lines joining the foci and the point of tangency. 
Let P be any point of the ellipse, through which 


the line P’PR is drawn so as to make the angles ? 
FPR, F'PP’, equal; then will P’PR be tangent at the 
point P. For, producing F’P to Q, and making #* Fig. 44 


PQ =PF, we have 
FP'+P'Q> FP+PQ=FP+PF. 

Now, if P’PR be not a tangent, let the second point, in which it 
cuts the curve, be P’, which we are at liberty to suppose, since P’ 
may be any point of P’PR; then the definition of the ellipse gives 

F’'P’ + PF = F’'P+ PF, which is less than F’P’+ P’Q; 

“ PF <PQ, and... 7 PPF< P’PQ, or FPR > QPR=F'PP, 
which is contrary to the hypothesis ; hence, so long as the 7 FPR 
= FPP’, the line P’PR cannot cut the ellipse, and is tangent to it. 
Q. E. D. 

Scholium. It is on this account that F, F’, are called the Foci of 
the ellipse ;_ since, from the principle of light and heat, that the 
angle of reflection is equal to the angle of incidence, if the curve 
were a polished metallic hoop, and a flame placed at F, the rays, 
reflected from all points of the ellipse, would pass through F’. 


Cor. 1. The tangents drawn through opposite extremities (220) 
of any diameter, are parallel. 


L-> 
[See (206), (99). ] IN 
<—Z 


Fig. 442. 


NORMAL TO ELLIPSE. "Spe 


Cor. 2. A parallelogram circumscribing an ellipse will be (221) 
formed by drawing tangents through the opposite extremities of 
any two diameters. 

Cor. 3. The tangents drawn through the extremities of (222) 


the axes are at right angles to them, and the circum- 
scribing figure becomes a rectangle. Cl, 
Fig. 443. 
Cor. 4. The Normal, or line drawn through the point (223) 
of tangency perpendicular to the tangent, bisects the an- 
gle embraced by the lines drawn from the point of tangency 
to the foci. Fig. 444. 


Cor.5, The axes are normal to the tangents drawn through (224) 
their extremities, 


PROPOSITION III. 


To find where the normal intersects the axis of abscissas. 


Let TX, =7 be the normal, intersecting the, axis T 
of X in X,; from (142) we have the proportion dt 
EF’ xX, F 


= " TF od C—f, bt , putting OX, = 2, Fig. 45. 
Sy PE es ty Ge 
Qu 


Cc 
=—_,.. Z,=—-«4u, in which, substituting the 
oo ea. a : f 6 


value of u, = Sa x, found under Prop. I., we have 
a 
c? F 
Xe, ha ake the point required. (225) 


e 
We observe that when c becomes = 0, z, [= iat x | 
a 


becomes = 0, and .*. the normal passes through the centre, O, as it 
ought to do, since the ellipse becomes then a circle described with 
the radius a. We have 


Subnormal = XX, = OX — OX, =2r—-7, (226) 
c? a* — c? b? T 
gah oie Lema 2 es z t 
» es 
We have (129), Fie. 459. ° 


120 EXERCISES, 


Subtangent[=XX,] Ordinate [TX] _, 
Ordinate [TX] — Subnormal [XX,] ’ 
(Ordinate)? TX? 


Subtangent = Subnormal XX, Se 
2 
y? 5 (ae a? — 72 


We observe here that the subtangent is independent of the value 
of 6b; therefore, 

Cor. If upon the same major axis, any number of ellipses (228) 
of different breadths, and also a circle be described, 
then their tangents, drawn to the same abscissa will 


intersect the axis of z in the same point. Fig. 453. 


EXERCISES. 


1°. Prove that if two points of a straight line glide along two 
other lines intersecting at right angles, any third point of the first 
line will describe an ellipse. 

2°. The equation of an ellipse referred to rectangular codér- 
dinates is 

9y? + 42? = 36. 

Required the distance from the origin to the point in which the 
normal cuts the axis of z, the abscissa of the point of tangency 
being z,=1. 

3°. Where does the normal in 2° cut the axis of y ? 

4°, Required the subtangent in 2°. 

5°. Required the length of tangent in 2° intercepted by the axes 
of codérdinates. 

6°. How far distant from the centre are the foci in 2°? 

7°. What is the eccentricity in 2° ? 

8°. Required the parameter in 2°, 

9°. It is required in 2° to transfer the origin to a point in the 
ellipse, the abscissa of which shall be z = 1°2, the new axes being 
parallel to the old. 

This will be done by substituting for y and z, y+ y and z+ 2, 
= x-+ 12, and observing that 

9y,? + 42,? = 36, or 9y,? + 4 + 12? = 36. 
10°. Given 9y? — 907+ 42? + 56x + 385 = 0, the equation of a 


HYPERBOLA, 121 


curve referred to rectangular codrdinates ; it is required to ascertain 
whether the curve be an ellipse. 

Substitute for y, y+ y, and for 7, 7+ 2z,; find the coéfficients of 
the first power of y and the first power of x in the new equation, and 
put these coéfficients separately = 0, from which deduce the values 
of y,, 2, The resulting equation will be found to be 9y? + 4x? = 36. 

11°. According to Sir John F. W. Herschel, the equatorial di- 
ameter of the earth is 7925‘648 miles, and its polar diameter 
7899170 miles. The situation of a place in north latitude is such 
that a perpendicular dropped upon the earth’s axis will intersect it 
at the distance of 2456 miles from the centre. Required the point 
in which the direction of a plumb line suspended at the place, will 
cut the axis of the earth; the meridian being regarded as an ellipse 
and the plumb as perpendicular to the surface of still water. 


SECTION THIRD. 
The Hyperbola. 


Definition. A Hyperbola is a plane curve described by the inter- 
section of two radii varying in such manner as to preserve in dif- 
ference the same constant quantity, while they revolve about two 
fixed points as centres. 

Ordering all things as for the equation of the el- x. 


lipse, except that the radii are to be denoted by w+ a, +5 ae 
u—d, we find Fig. 46 
a*y? — b?x*? = — a*b’, (229) 


for the equation of the hyperbola, 

The properties of the hyperbola are obviously analogous to those 
of the ellipse. The student will exercise himself in ascertaining 
them. 

It has been remarked that the equation of the ellipse, a*y? ++ b?z? 
= a*b*, becomes that of the circle, y? + 2? = a®, by making b =a; 
so the equation of the hyperbola becomes y? — z* = — a’, by put- 
ting b =a, an equation much resembling that of the circle ; hence, 
the curve which it represents, the Equilateral Hyperbola, pos- 
sesses properties analogous to those of the circle. 


SECTION FOURTH. 


The Parabola. 


Definition, The Parabola is a plane curve, such that any one of 
its points is equally distant from a fixed point and a line given in 
position, which line is denominated the Directriz. 


PROPOSITION I. 
To find the Equation of the Parabola. 


Take the directrix, D, for the axis of y, and for P 
the axis of x the perpendicular to D, drawn “i 
through the fixed point, F, whose distance from D yy 
we will indicate by p. Then, by the definition, Fig. 47. 


P representing any point of the curve, we have 
PAE ey es 
we y? = 2px — p’, (230) 
is an equation of the parabola. 
If in this equation we make y=0 for the purpose of finding 
where the curve cuts the axis of z, there results 
4 ar \GPE 
oe gat Fig. 479. 
Cor. 1. The Parabola cuts the axis of x midway be- (231) 
tween the fixed point F and the directrix, 


In order to transport the origin to this point, 
we have only to substitute in (230) for 2, 7-+49; 
doing which, there results 


Fig. 473. 
| y? = 2pe, (232) 
the equation of the parabola. 
Here we observe that as z increases y increases, and that with- 
out limit, and for every value of z there results two equal values of 
y; also, z does not admit of any minus value, since in that case 


y, [ =(—2)2], would be imaginary ; hence, 


TANGENT TO PARABOLA. 1% 


Cor. 2. The parabola opens indefinitely to the right in (233) 
two symmetrical branches, but, unlike the Hyperbola, has no 
branch on the left of the origin. 


PROPOSITION II. 


To draw a tangent to the Parabola. 

Let PP, be any curve, cut by the FP 
line P,PX,, intersecting the axis of 
xz in X,; then, denoting the codrdi- 
nates of the point P by y, z, and of 
P, by y+hk, c+h, we have 


Xs Xe My x 
Fig. 48. 


y ae 

subsecant XX, h’ et) 
Now it is obvious, that, if the point P, be made to approach the 

point P until the two coincide, the line P,PX, will cease to be a 

secant, and consequently become a tangent at the point = To 

effect this we have ee to make 

h diminish till h = 0,and .. k=0; 


indicating what the ratio © be- 


comes under this condition, by in- 


cluding = in brackets, we have 


y c- ay A 
subtangent X,X, _ & : (254) 


where the value fe is to be drawn from the equation of the 
curve. 


Applying this process to the parabola, we have 


y' = 2px — p’, 
and (ytky= ee +h) — 
(y+) —y° = 2p(z + h) — 2pz, 
or Qyk +k? = Wh; 


ae sales 


which substituted in (234,) tg 


es 
subtangent y’ 


hs 


124 SUBTANGENT AND SUBNORMAL, 


2 
subtangent = : (235) 
But we have in all cases [ fig. 482] 
subnormal >. y 
y ~ subtangent ” (208) 


for the parabola we find 


5, JRE 
subnormal X,X, = awe # =p (237) 
P 
Whence we have Y 
FX, = X,X,—X%,F=p—(p—2)=2; 2 Ey 
but FP =7z, .°. FX,= FP, 
7 Exe = FPR: xX F x, 


therefore, if we draw a line PX, parallel Fig. 483. 
to the axis of x, the angle X,PX,= X,PF. It follows that all rays 
of light or heat, or of sound, parallel to the axis of the parabola, 
will be collected in F.. Hence F is denominated the Focus of the 
parabola, 

Cor. 1. The points where the normal intersects the para- (238) 
bola and its axis, are equally distant from the focus, and the nor- 
mal is consequently equally inclined to the axis and to the radius 
vector, terminating in the same point of the curve ; 

Cor. 2. The tangent makes equal angles with the axis (239) 
and the line joining the point of tangency and the focus. 

In equation (232) the abscissa of the focus is 4p ; 

Parameter = 2y,_ 4, = 2p. (240) 

Scholium I. The method here employed for drawing a tangent 
to the parabola is obviously applicable to all curves ; and it is recom- 
mended to the student to make himself familiar with it by drawing 
tangents to the circle, ellipse and hyperbola, and to verify his re- 
sults by the properties already demonstrated in regard to these 
curves. 

Scholium LI, We must not pass unnoticed the remarkable symbol, 
by which we have readily arrived at important relations, Since in 


ea we have reduced both h and & to zero, it is natural to regard 


this expression as equivalent to 0° and this in itself, abstractly con- 


sidered, has no meaning at all, for to it we cannot attach any idea 


SCHOLIUM, 125 


independent of its origin. But to regard the symbol H as des- 


titute of signification, or not indeed as possessing an important 
one, would be to attribute to it an altogether erroneous interpreta- 
tion. In truth, it not only indicates a quantity, but that quantity as 
evolved, by a peculiar operation, from specific conditions, 


The symbol [z] signifies, 


1°. There are two quantities which are regarded as variable, y 
and xz. 

2°. y is regarded as depending upon z. 

3°. Increments [increases], k and h, are given to these vari- 
ables, y, z. 


4°, The ratio, ie of the increment of y to that of x is found. 

5°. That particular value, sel of this ratio is taken, which is 
obtained by diminishing h, and consequently k, to zero. 

It is also to be observed that the ratio, Ea will generally it- 
self be a variable quantity. Indeed, in this particular case of a 


tangent to the parabola, we have <1 pe _ , which may vary from 


_P _ — infinity to By Rab! Va = 0. In the next book we shall 
y= y = infinity 


give the symbol, [F<] , a name, and a further investigation. 


PROPOSITION III. 


If a curve be such that the distance of any point of (241) 
it toa point fixed in space shall bear a constant ratio, e, to the 
distance of the same point of the curve toa given line or Direc- 
trix,then will the curve be either the Ellipse, Hyperbola or Par- 
abola, according as the ratio, e, is less than, greater than, or 


equal to unity, [e $1]. 


126 GENERAL EQUATION, - 


Let the fixed line or directrix be the axis of y 
y, and the perpendicular to it drawn through the 
fixed point F, the axis of z, and let the distance a \z 
of F from the origin be denoted by d; there re- +—~—*—~.—> 
sults, Fig. 49. 


y? + (d— zx)? = 2? = ex’, since i =e, by hypothesis ; 


y? + (1 — e?)x? — 2dz + d* =0, which becomes at once 
y” = 2dzr — d’, 


the equation of the parabola, when e = 1, or (1—e?)=0. In order 
to make the term 2dz, affected by the first power of 
x, disappear, we will transport the origin to the 
right a distance = m, so that we shall have r-+m 
instead of z, 

*. y? + (l—e?) (c«#+ m)? — 2d(r + m) 4+ d?=0, 
or y?+(1l—e’)z?+[(1 —e?) « 2m—2dJzr+4+(1 Fig. 499. 

— e*)m® — 2dm + d? =0, 


from which, attributing such a value to m as to make the term af- 
fected by the first power of x disappear, we have 

(1 — e?) «2m — 2d=90, 
and y? + (1—e?) x? + (1 —e?) m?—2dm+ d?=0; 


whence, eliminating m, there results 
e2 
y’ + (L—e*) =, 


y" x 


e?d? a eed? 


I—e? (1—e*)? 


or 


2 2 

which is (1°) the equation of an ellipse _ + - =1, 
Ey had 
eo 
according as 1—e? is +, or —, thatis,e <1, ore > 1. Q.E. D. 

Scholium. It is to be observed that the Ellipse, Circle, Hyper- 
bola, and Parabola may be represented by the same general equa- 
tion, and are therefore to be regarded as nothing more than species 
of the same curve. 


2 
or (2°) the equation of a hyperbola, eet 


EXERCISES. 127 


EXERCISES, 


1°. From the top of a tower 48 feet high, a cannon ball is fired 
in a horizontal direction with a velocity of 1000 feet per second. 
Required the distance from the foot of the tower where the ball will 
strike the horizontal plane on which it stands ; no allowance being 
made for atmospheric resistance, and the vertical descent being in 
the times 1, 2, 3, é&c., seconds, 1? + (16-4), 2? « (1674), 3? + (1674), 
&c., feet. 
2°. To transport the origin to a point of the Parabola, the new 
axes being parallel to the old. . 
Let b, a, be the codrdinates of the new origin re- 
ferred to the old axes ; we have 
| (y +5)? = 2p(z + a); 
but b? = 2pa; 
y? + Why = pz, is the equation required. Fis: 50. 
3°, To Spee whether a board cut from a log next the roots 
without having been squared, may be regarded as an inverted par- 
abola. 
Let the middle line of the board be taken for the Y 
axis of z and the broader end for the axis of y; 
the preceding problem gives us 
(c — 9)? + 20(c — y) = pz, 
where there are three constants, b, c, p, to be de- ~ 
termined, one of which, c= half the width of the 
broader end, may be supposed known. We must therefore deter- 
mine the values of 5 and p from values of z and y taken in two dif- 
ferent places, and then see if 5 and p remain the same, or nearly 
the same, for measures taken throughout the length of the beard. 
4°, The length of a board, of the form given in 3°, is 8 feet, the 
ends are 4 and 2¢4 ft. broad, and the breadth of the middle is 3 ft. 
Required the equation of its edge, the axes of coérdinates being 
as in the last, Ans, (2—y)?-+ ‘7(2—y) = ‘152, 
or y? — 4°7y = ‘1dr — 544. 
5°. It is required to form a gauge by which to turn a parabolic 
mirror 18 inches in diameter, and having a focal distance of 10 
inches, measuring from the diameter. Required the depth of the 
mirror and its equation. 
Ans. Depth = 1‘7268. Equation, y? = 469072 « z. 


Fig. 51. 


128 EXERCISES. 


6°. Fold the lower corner of the left hand page, so as to make 
the area of the folded part constantly equal to a given square (a*) 
the side of which is a; and find the locus of the corner, or the 
curve in which it is constantly situated. 

If we take the position of the corner before folding for the origin 
of codrdinates and the two edges of the leaf for the axes, we shall 
find 

(y? + 2%)? =2 (2a)*yz, 
for the equation sought; from which it appears that the curve is 
symmetrical in reference to the axes or edges of the leaf, and that 
it begins and ends at the origin, since z =O gives y =O. 


oe 


sa wiht Yi) Wy ay me 3, |e alls hl Be 


ANALYTICAL FUNCTIONS, PLANE TRIGONOMETRY, 
AND SURVEYING. 


mes ee ie ‘Sesh mae commen as tho? \s ery: | enna te 
x a ie £ igh Ebadi dl Ste a ghvem wigdaie F 
ae | ie bir kalig eh pi ap ‘2 fitted ia dy ey ‘he, ial 
ae. : Bao, sharacgeh bi % oun estab Ct Vat ‘ 2. oe 
Flee ae fr onesona a serve sive 
| bapen Sion K san. 0. hes can, hia at Bin | sve he where, HE, 


caver bing jesiadan 
1 dct skh ie enh aa 


se ey Pop. ih ai 


. 
. 


ie we The PST bes Foam os tlie 


oA Bay att erica ete 1 aR 


anoose THAT | 


at Muon avets anrron’ saemretarh 
FA eine suena awa... rk, 


a 
| | semen A cerns ceeatiaats 
Ne nee sey cere a Bar Te eT eel sgn 


BOOK FIRST. 


ANALYTICAL FUNCTIONS. 


SECTION FIRST. 


Primitive and Derivative Functions of the Form a 


Definition 1. When one quantity depends upon another, so that 
the first varies constantly by a continuous change of value in the 
second, the first is said to be a Function of the second. The first 
is also called the function and the second the independent variable. 

Thus, in the equation of the parabola, y? =2pz, y is a function 
of x, or y is the function and z the variable. 


Def. 2. An increasing function is one which increases when the 
variable increases, and a decreasing function is such that it de- 
creases when an increment is given to the independent variable, 


Thus, in the equation y*? =2pz, y is an increasing function of z, 
. , ‘ bY. 2 
but in the equation of the ellipse, y? = a (a? —x*), y is a decreas- 


ing function. 
Def. 3. An explicit function is one in which the value of the 
function is developed or expressed; an implicit function is one in 


which the value of the function is implied. Thus y = (2p)? . 2, 
is an explicit function; and y is an implicit function of z, or x an 
implicit function of y, in the equation a*y* + 6?2? = ab’, 

As it is useful to denote quantities and their combinations by let- 
ters and signs, so it will be found advantageous to indicate differ- 
ent functions by appropriate symbols. The letters f, F; fi, Fi, 
&c.; f', F’, &c., as well as others derived from the Greek alpha- 
bet, such as ¢ (phi), ¥ (psi), are employed for this purpose. Thus 


132 DERIVATIVE. 


fz signifies, not that f is multiplied into z, but that some operation 
is to be performed upon z, such as squaring it, taking its square 
root, &c., in which case we have fx = 2’, or fr = ./z, &e.; fx will 
signify a different function or operation from f, z, and the distinc- 
tion in reading will be, the f function of z, the f sub one function 
of x. Further, the small letters may be used to express known re- 
lations, while the capitals denote unknown functions. 

The letter h will always be employed to denote the Increment or 
increase of the independent variable z, while & will as constantly 
indicate the consequent increment of the function y. 

Def. 4. The Derivative of a function, or the Derived Function, 
or simply the DerivarTive, is the function obtained by taking that 
particular value of the ratio of the increment of the dependent 
variable to that of the independent when the latter reduces to 
zero, 


Thus, if pee foes, 
then* y+k=(e-+h)'=2"+2ch- +H 
wae k=22rh+h?; 
k 
ae =2r+h 9 


and .°. be | = 2z, the value of the ratio a whenh=Oand ..k=9, « 


indicated by the brackets [ ], is the derivative of the function 2°. 
And, as y is put for the function 2’, so it is natural to indicate the 
derived function 2z by y’, which will therefore be read the deriva- 
tive of y —, and we shall write 


y-[E]-m 


The letter f’ will be employed for the same purpose, and f’z will 
be read the derived function of z, As a further illustration, let us 
take the function 

y = fr =azr>+b, 
then* y+ =f(2-+h) = a(a+h)? + b = a(x? + 3827h + Brh?+h')+. 
k=f(x+h) —fe = a(32*h + 3rh?-+ A’), 


k rth)—fxr 
and - uiticid we = a(32z*°+ 32h -+ h?) ; 
* Increasing x by 4, and, consequently, y by &, or changing x into x-++-A and 


y into y +k, or, y depending on x by a constant law, y+ is the same function 
of x-++-h that y is of x. 


DERIVATIVE, 133 


yay Ea = pao =ae Sz’. 
It is obvious that y’, f’z, Ea : ey , are but different 


expressions for the same thing. Observe further that y’, being 
equal to a « 32’, is also a function of x and its derivative may be 
found ; so that from 

y =f = an dz", 
we have Yah 2A aeZe We, 
which is the derivative of y’, or the derivative of the derivative of 
y, or simply the second derivative of y. In like manner the third 
derivative of y would be indicated by y'” or fx, the fourth by y” 
or f*, &c. 


PROPOSITION I. 


To find the derivative of any function capable of being devel- 
oped in integral additive powers of the variable. 

Let the function be denoted by 

y=fr=aqtazr'+a zr? +az7?4+..ta2°+... (242) 

Increasing z by h and the function by k, we have 
ytk=f(x +h) =a,4+a,(c+h)!+a,(r+h)? +... + a,(c+h)"+ o05 

. k=f(e+h)—fe=af(et+h)'—2z']+af[(x+h)?—27]+... 

+a.[(z7+h)*—2z"]+..., 
a. Peal oo ae 

Ae ae a ee 


q h 
dif AER irk ay 
(f+ h)isaeh oh. 
but. aaa 
OT eg ee erat ag tnt b 
Seuvn-at 2p 2 3 
——- bast slp B32? + (3r-Lhyh, 


ta 
de Siar will be found=4z*-+ PA, where P depends 
upon z and h; and we are led to infer that 


Ge = ng Xh, 


134 DERIVATIVE. 


where X is such a function of xz and h that Xh shall disappear when 
h is diminished to zero. But to prove this let us put 
xrth=a};x=b, and ..h=a—); 


Ra have (e+h)'—2'*a—b' (cx—h)’?—2z’ | a’—5? 
oe Lem 


a—b’ h a—b’ 
(x+h)?— x8 Pahari" ee (x +h)" ~ 2" _ a" —~ be 
h Oe Bi h a—b’ 
so that we fall upon the examples under (16), and we have only to 
solve the problem whether a" — 0” is divisible by a—b. 
In order to this we execute the multiplication, 
(¢— b) (a"-' + a? b+ a" b2 + ar 53 degree £ ab 5s), 
and we find the product to be 
a” +a" b+ a” b? + a®$b3+ a? bt +... + a?b"? + ab 
—a"*b—a?* b? a 63 — a" Bt — .., —a*b"* — ab" — 3’, 
which is equal to a* — 8"; 
a” — b” 
air es 


= qr + ah fe a’ $2 +...+a7b"?+ab""+6"", (243) 


Hence, the difference of the same powers of any two quantities 
is divisible by the difference of the quantities themselves, and the 
quotient is homogeneous and one degree lower than the power, the 
leading quantity descending one degree each term and the follow- 
ing ascending by the same law. 

If in (243) we now replace a and b by their values z +h and 2g, 
we find 


} 2 A man 
nie ar al 2 <= (2 hy (wth) te + (ah) ea? +... 2; 


whence the fourth equation becomes 


k z+h) — 
FEE «a, 4a, (Qe +h) +05 (80?-+8zh-+ M) +o 
+a, [(a + hy + (x + hye + (2 + hye? 4... + 2]; 
therefore, making h = 0 and observing that there are n terms in the 
expression (x + h)""'+..., which all reduce to x”, we get 


y =f'x = ka = Peto] 2 at a8 Qa-ba, © Bx® (244) 


+a,+ 47+ ,..+a,+nz"'; which is the derivative sought. 
We perceive that (244) is derived from (242) by multiplying by 
the exponents of z in the several terms and decreasing them by 


DERIVATIVE OF Az’. 135 


unity ; and hence that the term a, independent of z, or, which is 
the same thing, the term a, + 2°, disappears in (244) ; so that, if 
we were to pass from (244) back to (242), it would be necessary 
to introduce a term independent of z, or to add a constant quanti- 
ty, which may be represented by C. 


PROPOSITION II. 


To find the Derivative of any real power of a variable. 
Let y = fr = Ar’ be the function, a being any whatever, whole 
or fractional, plus or minus, but not imaginary, not of the form 


a=-/—c. Suppose, in the first place, that a is fractional and 
additive, 


™m 
or Ce ee 


then y= fr= Aa" = A(x") . 
Now, if we substitute z for 27, we get 
y =f.z = Az”, and z=fyr =a". 
Suppose then that, while z is increased by h, z receives the in- 


crement 7, and, as a consequence, that y is augmented by k; we 
have 


y+k= A(z-+ 7)”, and z-+i=(c+h), or (z+7)"=zr+h; 
w k=A(z+i)™— Az”, and (z+ 7)"— 27=h; 


k.. (z-+%)"— 2 i z x 1 
B) gem [ ase aaaP ES NG leneeaact (z+ 7)" = 2"’ 
a 
haley RM, ” Lacie mdi 1 ; 
; hi ‘4 i (z+i)"—2"’ 
| t 
k k al wey he Ee eee 
. (E)- EEE] earnest ghee Be 
eA (Se) we rz", 
n n 
™m =i 
Or Yynse= Yong 29 Miape= Ave @ 


* y’y-fe, read, the derivative of y, y being a function of z. 


136 DERIVATIVE OF Az’. 


Here it is worthy of remark that we have found the derivative 
of y a function of x, by taking the product of the derivative of y a 
function of z, and that of z a function of x We observe the 
same rule, then, for the derivative, whether the exponent be frac- 
tional or integral, we multiply by the exponent, and diminish it by 
unity. 

Let a be now taken subtractive, and either integral or fractional, 
or let 

a=—T, » 
where 7 is either a whole number or a fraction ; we have 
y= Ar*= Ar; 

* Ytk= A(zr+h), 
and k= A(x+h)"— Az”; 

k as (z+ h)7 — Ghat er (c+h)7—a27 (e¢#+hyr’ 


h h h "(et hyn 
ace (2- hye" — a2 -+- hy iy x2’ —(x#+hy 
(2 +h)'2’h fe (c+ h)a2"h 
(cx +h)— 2” 
h 
A EE 


k r—1 
Bi Sh | =| =_— A e ae =A e —rz¢ 7, 


the derivative sought. Hence, 


RULE. 


To find the Derivative of a variable, affected by any (245) 
exponent whatever that is not imaginary; multiply the variable 
by its exponent, and decrease that exponent by unity. 


Thus, the derivative of z? is 2x”, of x* is 32°, of x is 4z°; of 


—1 —4+ Be Pe 2 
hd 3 3 ° 4 
zi isfo. = f2,°=F+—, of g istsa” =fex", of a is 
x3 
1 : 
210g Tee Os of Az* is Aar*"; of ris] egiIt=1-. 7° 


=i Le 1: uf Ape ae) Lie As De 4 Oe 
& 


DEVELOPED FUNCTIONS. 137 


PROPOSITION III. 


The Derivative of any function, capable of being devel- (246) 
oped in real powers of the variable, will be found by multiply- 
ing in each term by the exponent, and decreasing it one. 

ConverRsELy: if a derived function be developed in powers of 
the variable, we shall return to the original function, by increas- 
ing the exponents by unity, and dividing by the exponents thus 
augmented, taking care to add a constant. 

Thus, let y be such a function of z, or depend upon x in such 
way, that we have 


y =fe = Ar*+ Ba’+ Czr°+..., (247) 
where A, a; B,b; C,c; &c., may be any real quantities, + or 
—, whole or fractional, we have (245) 


y =f'r= [+] = Adar + Bo 4Coa +... (248) 


We observe that if a= 0, or there be a constant term, Az’ =A, 
in (247), it will disappear in (248), since Aaz*" then becomes 


1 
A+0+—=0; therefore, in passing back from (248) to (247) we 


must add aconstant for that which may have disappeared. We 
shall see how this constant will be determined in any particular 
case by the nature of the problem, 

To illustrate, suppose we have found the derivative 


y = 3x — f05 +7, 
returning to the function (246) we obtain 
y =8r? —Z- bys + 7x + constant. 
Whenever, then, we can find the derived function developed in 
powers of the variable, there will be no difficulty in ascertaining 
the primitive function. 


PROPOSITION IV. 


The Derivatives of equal functions, depending upon (249) 
the same variable, are themselves equal. 

This is manifest from the nature of the operation; thus, if we 
have any two functions of zx, such that 


Ex = $2; 


138 © SCHOLIUM. 


whatever may be the value of z, then are the functions equal when 
zx is increased by h, or when z becomes x -+-h, and we have 
F(x+h)=f(x+h), 
from which subtracting the first equation, there results 
F(z +h) — Fr =f(z +h) —fa, 
which, divided by h, gives 
F(x+h)— Fr _f(x+h)—fe. 
h r h ; 


and this equality, having been obtained independently of any sup- 
position in regard to the magnitude of A, is true for all values of h, 
and therefore true when A becomes less than any assignable quan- 
tity, or when / is diminished to an equality with zero. 
Therefore, Relea aed mes de oa = [ Age ==) : 
h h 
or Se MS Be 

Thus, if 

A +A,e z'+A,-27?+..=aq+a,¢zr'+a,¢ r?+..., 
then shall we find 
A, +A, ¢2r7+...=a,+a,¢ 2r+.... 

We observe that if the last equation be multiplied by x, and sub- 

tracted from the preceding, we obtain 
A,— A, + 2? — A, « 27? — ... = d)— Gye ©? — Og e AT? —..., 

from which the first power of zx is eliminated ; and, in like manner, 
we might eliminate in succession the terms affected by x’, 7°, &c., 
to the last, if the series were finite, and to a term less than any as- 
signable quantity, if infinite and convergent, that is, if z were less 
than unity; hence we should obtain A, =a,, and, pursuing the 
same process, A, = a,, A, = a, and so on, which is in accordance 
with (65). 

Scholium I, The student will not embarrass himself with unne- 
cessary difficulty in reference to the symbol 


’ [1 or its equivalent Sea ’ 


by supposing it to signify nothing more than is indicated by re in 


itself considered, It is true that the numerator as well as the de- 
° ° k ° e . 
nominator in loge is =Q; but we are not to infer from this, 


either that the quotient, [=| ,1s equal to unity, or that one 0 can 


SCHOLIUM. 139 


have a different value from another 0, in order to make the quotient, 


0 
@ ’ represent different quantities ; but that [+] is simply asym- 


BoL of a determinate operation, whereby we have derived one 
quantity from another. We are not, therefore to.separate the & 
from the A as if they were determinate quantities, and as such 
could be managed separately, nor even to remove the brackets ; 


the whole expression, [+]. let it be remembered, is one indivi- 


sible symbol, the sign of a process, it does not so much signify 
what the result is as how it is obtained. In any function, de- 
pending upon z, we give to x an increment h, we subtract the un- 
augmented function away, we divide by the increment h, and, in 
the quotient, eliminate 4 by reducing h to 0. Nor is this a singu- 
lar case; all algebra is but a system of symbols, a mathematical 
language, telling what is to be done, and showing how results 
are obtained. Thus, ab, a: b, a’, /a, are not simply the re- 
presentatives of quantities, but indicate the operations, multiplica- 
tion, division, involution, evolution, whereby certain quantities are 
obtained. . | 

Scholium II. It is obvious that the derived functions will differ 
from each other according as the primitive functions, from which 
they are derived, are different. 


SECTION SECOND. 


The Binomial, Logarithmic, Interpolating, and Exponential 
Theorems. 


PROPOSITION I. 


It is required to develope (a+ x)" in powers of x, n being any 
real quantity, plus or minus, integral or fractional. 

First, suppose n to be plus and integral, that is, any of the num- 
bers, 1, 2, 3,4, 5,.... It is manifest that the development of 
(a-+ x)" can contain no other than plus integral powers of x, and 
must contain these, since the multiplication of whole positive 


140 BINOMIAL THEOREM. 


powers by whole positive powers gives whole positive powers ; 
thus, 
(a+ 2)? =(a+2)(a+27)=@?+2ar4+ 2’, 
(a+ 2)? =(a+ 2) (a?+42ar+ 2’), &e. ; 
hence, (a+ x)" must = A,+ A, + z'+ A, x*+ Azo 7+... 
+A,*2"+..., (a) - 
arranging x in ascending powers, and denoting any term by A, « 2”. 
In order to determine the coéfficients A,, A,, As, ..., A, of the 
several powers of z, it will be necessary to take the derivative of 
(a); therefore, to find the derivative of the first member (a+ 2)’, 
put 
y=(a+2)"=2", and .. z=at+z; 
ytk=(2+72), and zt+i=a+e2t+h; 
_@+ipne 
a 


t 
and =1; 


+. agri a on 


Yy=fe = tf |= E 4, [= | ant anata, 


whence the same rule holds for the derivative of (a-+ x)" as for 2’, 
substituting a+ 2 for z. 
The derivative of equation (a) gives us then (249) the equality 
n(a+ar)""= A,+ A, + 2r+ A, + 32?+ Aye 47? +... 
+A, * mz”""+ ...3 (6) 
Taking the derivative of (b), we have 
n(n —1)(a+ar)’?= A, + 2+14A, +36 27+ A, eo 40 Bz? 
+..+ A, + m(m —1)r"?-+...3 (c) 
the third derivative gives 
n(n —1)(n— 2) (a+ 2r)"%=A,+-3-2e-14+A, 2463 Bz 
+..+A, © m(m —1) (m—2)z"*+...5 (d) 
the fourth derivative will be found 
n(n —1) (n — 2) (n— 3) (@4+2)"%= A, + 4-3-2-1+4,,. 
+ A,, « m(m — 1) (m— 2) (m—3)2""* +... 5 (e) 
therefore, observing the law of derivation in (a), (b), (c), (d), (e) 
-., and proceeding to the mth derivative, we obtain _ 
n(n — 1)(n—2) (n—8) (n—4) ... (n—m+1)(a+-2)"™ 
= A,, + m(m — 1) (m— 2) (m—3)...¢3eBelt+..., (1) 


BINOMIAL THEOREM. 141 


From the last derivative we observe that, if 7 be a whole addi- 
tive number, m can never exceed n; for if m were =n-+1, 
2—m-—+1 would be =0, and the (n-+1)th derivative, and con- 
sequently all after it, would disappear. his is as it should be, 
since it is obvious that the nth power of (a+ 2) cannot give any 
term in the development of a higher degree than n. 

As the coéfficients Ay, A,, Ag .., Any +, are independent of z, 
we may make z =0 in (a), (0), (c), (d), (e), -.., (2), ..., and we ob- 
tain 

a" = A,, na" = A,, n(n — 1)a** = A, oe, 

n(n —1)(n—2)a"* =A, + 3-21, 

n(n —1) (n — 2) (n—3)"4=A, + 4-3-2 1,... 

n(n — 1) (x — 2) (n — 38) (n —4) ... (2 —m- 1)a"™ 

= A ® m{m —1)(m — 2)... 3 e¢2el, 

Substituting the values of Ay, A,, A..., A,, drawn from these 

€quations in (a), we obtain 


Newrton’s Binom1aL THEOREM. 


{atzr)*=a"+na™" + x+n(n—I1)a"* ss (250) 
+ n(n —1)(n—2)a"*e aoe aby 
+ n(n—1) (n—2) (n—3) ... © (n—m-+1) a « sat : 


Since the same rules of addition, subtraction, multiplication, di- 
vision, and derivation are applicable, whether the exponents be 
plus or minus, integra] or fractional, we might infer that (250) 
would hold in all cases, whatever might be the character of n. 
But to prove it observe that if m be any real quantity [not imagin- 
ary |, there can be no other than real powers of z in the develop- 


é 
ment/such a mo, ily A, DiS Be egos vole Vike Uy, Blea. 
Therefore the form 


(a+2)"= Ayer’ pAyeat tif Ae oF tif A, ea 
Fi 


+ A_,¢ + AL, oot. 
must at least be sufficiently general, Suppose now that the minus 
powers are arranged in a descending order, that is, that r>s, 
s>t..; .. multiplying by z*, we have 


142 BINOMIAL THEOREM. 


(a-+eyea=(Ab Apert. t Ae eae.) a" 


Seaside Me of BBY hp eo 
where it is obvious that all the powers of z are plus; .*. making 
z = 0, there results 
0=0+4 A_,+0+..., .. A_L.=0; 

and in the same way it may be shown that A_,=0, A_,=0,...; 
that is, the expansion contains no minus exponents, Rejecting the 
minus powers and taking the mth derivative, we have 

n(n —1)(n—2)...¢(n—m+1) (a+ 2)""= A, + m(m—1) 

(m—2)...e3e2e1+A,.,°(m+1)m(m—1)... 27+... 


$A S(4-1) (4-2) ..+(S—m4ijor 4... 


The order of this derivative may be any whatever, if nm be minus 
or fractional, since in this case the coéfficient n — m-+ 1 can never 
become = 0, m being a whole additive number; we can therefore 


take m so great that the exponent Ar m shall be minus, (+ —m 
can never = 0, m being = 1, 2, 3,...), But 
n(n —1)...° (w»—m-+1) (a+2)" 

cannot have a minus exponent in its development, for the same 
reason that (a+ 2)" has none. It follows that the development 
of (a+ x)” will not be affected by any fractional power of x, what- 
ever n may be; (250) is therefore true for all the real values of n, 
since for all such values the coéfficients of the expansion have been 
determined. It appears, however, that if n be either minus or frac- 
tional, the series will never terminate, 

As the form of (250) is independent of the value of a or x, we 
may change z into — x, doing which and observing that the terms 
affected by the odd powers of z will be minus, we obtain 


(a—z)"=a"—na""-r+n(n—1) Z n(n —1)(n=2) 


v3 
eon We aa ne (251) 


If in (250) and (251) we make a=1 and x=1, and remember 
that all the powers of 1 are 1, we get 
Feng n(n—1) , n(x—1) (n—2) 
=ltat 5) iit e ae Daytime 
n(n—1) n(n—1)(n—2) 


and Valea en a cet ee ee 


BINOMIAL EXPANSIONS, 143 


Thus it appears (1°) that the sum of the coéfficients of any bi- 
nomial power is equal to 2 raised to the same power, and (2°) that 
the sum of the coéfficients of any binomial power taken alternately 
plus and minus is equal to zero. This furnishes a convenient 
method of verifying an involution. For example, 

(a+z)'=1lea+lezr, 14+1=2=2', and+1—1=0; 

(a+z2)*= a? + 2ar+ 27, 14+241=4=2', and+1—2+41=0; 

(a+ 7)? =a?+3a*%r+ 3axr? + 2%, 14+-34341=8=2', 
and 1—3+3—1=0. . 


Expand (a+ z)*, (a+ 2)°, (a+ 2)§, (a+ 2)’, by aid of (250), 


and (a—z)', (a—z)5, (a—z)*, (a—z)’, by aid of (251), and 
verify. 
If in (250) and (251), we make n = MS , there results 


ox thawte cesar arith in altage 1h. gel) 2° 


2Qr—1 38r—L 
P.O. RF. Zare rT? «23a * 
og 0 Pee 4 
wl 1) (2r 1) Ors 1) Hb ft af Bile (252) 


1 1 
and (a—2r)7=ar—-—_{ — 5,5, -—- >a —s (2888) 
rar TeeQar Te W3e4ar 


Making r = 2, 3, in (252), (253), we get 


a a 2 3 
(a-+ajtaah ++ —_* 4 
Qa” 2? « Qa*, BeBe Ba? 
3s 524 
pe ein ey tg, (254) 
2.2.3 4a’ 
a pit 2 3 
(o— x) =< gag nae tah teh es (255) 
Qa* 22a" 2.2. Ba? 
5 Gee Qr 2. 52° 
(a+2)? =a + —-— 4 
a> 32. Qa® 3? «2+ 3a° 
2.5 8rt 


nO seamen Xda Pe (256) 
31-23. 4a° 


144 BINOMIAL EXPANSIONS. 


ee we ar | (257) 
3a*° 3?» 2a° 
In (254) "RB a=c*, we find 


ee 323 
Q 7 ee ee 
(c +2)" =e+ 5 Bo 2. 2e8 + 33.29.30 
3-524 


ag Sct sn oat (258) 
from which, substituting z? for x, we obtain 
4 OS Zz 326 
2 2) 4. — — ——__—. ++ —_____ 
(oT 2) =O as oF. ac8 T BF. Bee 
3° 52° 
TH. 2.8 cas ears ow) 
changing z into — z in (258), 
ct— zr ew ard ieee 80 Ee a 
( ‘ 2c ° 2c? 02+ 3c? 
3° 5x4 
TRB. Sada ert (260) 
changing z into 2? 
L 2? 2 326 
Do ee elet ean Mee opt eee ak BA as Tee 
er ne me 2B a Bee. « So BeBe et 


The student may operate like changes upon (256), and investi- 
gate analogous forms for higher roots, as the 4th, 5th, &c. These 
expansions may be employed for the extraction of the roots of 
numbers. Thus, let the square root of 101 be required. We 
have (258) 


4101? = (10? 4.1)? =104+4 — sooo + teat Trev o000D 
= 10°0509381211. 


Find //102, 103, /104; ./99, ./98, /97, */1001, &c. 
We may put (250) under a new form, for dividing by a”, there 
results, 


(1+) =14n0-Z4n(n—1). _ 


Qa? 
+ n(n —1) (n—2) +5 tans (262) 
or (1 + Heed Att Ws 2 


4 


? x 
+..., putting v= cy (262) 


n(n —1)(n —2)+ 5-3 


LOGARITHMS, 145 


LOGARITHMS, 


Definition, Let a’ =z, (263) 
then is y denominated the Logarithm of x. The constant a is 
called the base of the system. 


PROPOSITION II. 


The Logarithm of a product, consisting of several fac- (264) 
tors, is equal to the sum of the Logarithms of those factors. 


For, let a= 2, ©, 15; + > Y=Yis Yor Y29 ove 3 


then (263) ai = 2; 
2 = Lay 
a3 = Do, 
&c., &c. 
Mirna@als Wie woiesati iar Yeti = 2) Morg elegy ’st 5 
or® L(x, + 22° 23..)=(Yy:tyw+tys +...) = Lr,+ Lr, 
+ Dr,+... Q. E. D. 


Cor. 1. The logarithm of the nth power of any number (265) 
is equal to n times the logarithm of the number itself. 


For, making z,= 7,=2;=..., we have 
L(z,+ 2, + 2, ¢...[n]) = Lr,+ Lr,+ Lr,+..., 
or, Lae \enLz,. (265) 


Cor. 2. The logarithm of the nth root of any number, (266) 
is equal to the nth part of the logarithm of the number itself. 

Cor. 3. The logarithm of a fraction is equal to the loga- (267) 
rithm of the numerator diminished by the logarithm of the denom- 
inator. 

For, if we divide a“ = x, 


by a2 = Loy 
zx 
we have ai~%=-'; 
Z2 
Lr; 
or L "e. =Y,— Y= Lr, — Lr. 
2 


Scholium, We perceive that addition of logarithms corresponds 
to multiplication of numbers, subtraction to division, multiplication 
to involution, and division to evolution. We have, then, only to 
possess a “ Table of Logarithms,’’ calculated to a given base, 


* I, logarithm of. 
10 


146 LOGARITHMIC DEVELOPMENT. , 


say a = 10, in order to perform numerical operation with remark- 
able facility. ‘Thus, to obtain the cube root of 2, nothing more 
would be necessary than to take the logarithm of 2, divide by 3, 
and seek the corresponding number from the table. 


Again, if 5 
then (265) L3 = L(2*) = zL2; 
Ey ae? 
AS 


whereby we are enabled to solve, with the utmost facility, a numer- 
ical equation in which the exponent is the unknown quantity. 


PROPOSITION III. 


A number being given, it is required to find a form by which 
we may calculate its logarithm. 


Since the relation a’ = x, gives y a function of z, let us endeavor 
to expand y in terms of xz. To this end we proceed to determine 
the derivative of y. 

Changing z into x+h and y into y+ 4, we have 


a’*F¥=er+h; 
subtracting Gok at, 
there results a ea'—a=h, 
or h=a"(a'—1); 


in which, substituting z for a’ and 1+-d for a, in order to subject 


a‘ to the influence of the Binomial Theorem, we have (262) 


2 
h=x[(1+0/'}—1J=2[1+hb+k(k—1) + = 
5° 
3.3 +..—1]; 
dividing & by each member of this equation, observing that 1,—1, 
cancel each other, and that & then becomes a factor common to the 


numerator and denominator of the second fraction, there results 
k 1 


¥ [b+ Ae a TS as i, 

sy tN R—2%) + a at 

from which, observing k =0 when h becomes = 0, we have 

; of & 1 1 

f= ]= pa Se, (0208) 
[s-sta—gtin [2 


4+ k(k —1) (k—2) » 


LOGARITHMIC DEVELOPMENT, 147 


putting M= 3 - : ; (269) 
5 pa 
POMS MAin grad) 
a 1 
Reece CST) CEs 
(a Lice 2 + 3 st) 4 +, ~~ 9 eve 


Applying the rule of (246), in order to return from the deriva- 
tive (268) to the primitive function, we fall upon the equation 

—1+-1 
—1+1 
from which z disappears, and nothing can be inferred—save that 
the logarithm of a number cannot be developed in terms of the 
number simply. 

But if we make y the logarithm of 1 + z instead of x, and repeat 
the above operation, from 

aw=l14+2, 

we obtain y =M(1+2z)"', 


or (262), y =M[14(-Io+(-I)(-1-)S4(-1)(-1-1) 


M 
y=M- + constant = my + constant, 


3 
= gt |= Mtl —z+4+2?—27'?+71—27'4+,-,...]; 
from which, returning to the function, we get 

y = M[z —42?+42? —42t + 427° —tr'+, —,...] + constant. 

In order to determine the constant, we observe that x and y van- 
ish together, since y=0 in a¥=1-+2, gives 14+ 2=a° =1, and 
.. £=0; therefore, substituting these corresponding values of y 
and z, we get 

0=M-0- constant, .. constant =0, and 

L(+ 2)=y=M[e — 42° + 40°40! 4 42°20 +,—,...}, (270) 
a logarithmic series, in which the logarithm of any number 1+ 2 
is expressed in terms of a number less by unity, z. 

The constant, M, depending upon the base, a, (269), is denomi- 
nated the Modulus of the system. Taking a different base, a, we 
obtain a new modulus, 


(-1-2). 


ip th | 


(ce) — Shy mes 


148 MODULI COMPARED, 


and the logarithm of 1+ z, derived from dy =1-+ 2, becomes (270) 
L{1 + x)= M[ zr — 32° + 32° — Fat +, —, J; 
by which, dividing (270), there results 
L(l+2)_ 
L(1+ 2) =n} 

Cor. 1. The moduli in different systems are to each other (271) 
as the logarithms of any given number in those systems, 

If we make the modulus equal to unity, we have the logarithms 
employed by Lord Napier, a Scottish nobleman, who was the in- 
ventor of this admirable system of numbers; and, if we assume 
10 for the base, we have the logarithms in common use, or those 
of Briggs. It is customary to indicate a Napierian logarithm by a . 
small 7, and a common logarithm by the abbreviation log. Adopt- 
ing this notation, and observing that 10”.=1-+ <2 gives 10'= 10, 
or log. 10 =1, we have (271) 

log. 10 M,_1 1 272) 
10 © jap ee Oe 

We have, then, only to make the modulus one in (270), and 
thereby calculate the Napierian logarithm of 10, in order to deter- 
mine the modulus, M,_,,, of the common system; but, by a little 
artifice, we may convert this series into one more rapidly converg- 
ent, and therefore better adapted to our purpose. 

In (270) changing z into —z, we have 

L(1 — z)= M(— z— 42? —47r?’ —...), 
which, subtracted from (270), gives 
Li+2)— L1 —z) =2M(2+42?+40'4+...), (278) 


or (267), L(; +5) = QM(x +409 442° +427 +...).. (274) 
And this series will be expressed more conveniently by putting, as 


Borda has done, 


1t+r m —n 
Rs » and .° ; whence 


Lae tn en m+n 


(2) =m ( (52) +22) H(SSa) +=]. em 


Making M=1, m=2, n=1, in order to calculate the Napierian 
logarithm of 2, we find 
U4) =22=2[4 +3(4)? +4(4)? +]; 


; hence, 


COMPUTATION OF LOGARITHMS. 149 


where we have only to divide successively by 9[= 3*] and the odd 
numbers 3, 5, 7, ..., as follows : 

3/2+00000000 
‘66666667 
‘07407407 
00823045 


1|‘66666667 
‘02469136 
‘00164609 
00091449} '7/:00013064 
‘00010161} 9)‘00001129 
0000112911 /*00000103 
‘00000125/13)|‘00000009 
00000014) 15/00000001 

-. 12 = 69314718 


Having thus obtained the Napierian logarithm of 2, /44=1(2 « 2) 
= 12+12=212= rene and putting m=5, n =4, 


~jF Ol 


CO moomoo oO OD Ol 


1=24+4. att se toa) = §22314354 ; 


9 


Operation. 


but bs BH vi M 
SI) 222222221 1 22222222 
sist! ie iy ee 1} 274348/3| 91449 

a | ; 
110 = 1(2+5) =12-4 15 3 we ; 
— 2:30258508 : mai ee 
vidtaaiaa 22314354 

(272), Mi, 19 = 75 = 043420448 


ce 


m—n\' m—n\3 

or log. m=log, n +-o-s6s59806[ (7 +) 4+3(e— a) +o]. 

Suppose, now, it were our object to construct a table extending 
from 1 to 10000, we should commence with 100; since, in calculat- 
ing up to 10000= 100 - 100 we should fall upon 200, 300, 400, 
&c.,=2 + 100, 3 - 100, 4 - 100, &c., whereby the logarithms of 
2, 3, 4, &c., would be determined, 

Since log. 100 = log. (10?) = 2log. 1O=2+-1=2, 


log. 101 = log. 100 ++ 0186858906 (95 : ial 


201 
0'86858896 
= 24 = 210043214 ; 


. (278), log. (= )- o-seasssoe| (7 — 


150 COMPUTATION OF LOGARITHMS. 


‘86858896 


log. 102 = log. 101 + ——-— = 2:0086002 ; 
log. 103 = log, 102+ Sai = 20128372 ; 
-&e &e. > &e. 


After two consecutive Jogarithms have been obtained, the opera- 
tion may be shortened; for, putting m =n-+1, we have 
‘86858896 
log. (n a 1) = log. n oe “Qn+1” 
‘86858896 | 
W+3 ’ 


1°73717793 
log. (n+ 2) = 2log. (n-+- 1) — log. n — ric eae pL (277) 


log. (n+ 2) =log. (n+ 1)+ 


Observe also that 4(2n + 3) is the difference of two consecutive 
divisors, 4(n + 1)? — 1, 4(n + 2)? — 1. 


‘ 1°7371779 
Making n= 100, log. 102 = 2log. 101 —log. 100 er Bae T Tam & 
Operation, 
10001 
+ 200 + 40086428 
oe kas — 20000000 
10201 Es 426 
4 1 one. 
—_—— .. log, 102 = 20086002 
40803) 1°7371779(426 
1°7371779 
r= 101, oe log. 163 = 2log. 102 — log. 101 ~ 40803 + 4(20243) . 
Operation. 
4:0172004 
40803 4 
— 20043214 
820 : 
eee ie _ 417 


41623) 1°7371779(417 ——— 
( . log, 103 = 20128373 


The student should continue the computation. 

Before terminating this problem, we will make one other trans- 
formation whereby the calculation of logarithms will be rendered 
far more rapid. 


COMPUTATION OF LOGARITHMS, 15] 


For small numbers, Borda puts* 
m=(p—1)*(p-+2) = p?— 3p+-2, 
n= (p+ 1)*(p—2) = p?—3p—2; 
a" m—n =A, 
and m+n = 2p?—6p; 
therefore, substituting in (275) and making M =1, we have 


(Frytp=3))*L(ams,) + pap) +4(pcq,) +] 
or 2U(p—1) + (p-+2)—2U(p +1) —U(p —2) 


is ic : =) Has) + | 


Making p = 5, 6, 7, 8, successively, we obtain 
212 — 313 +17 = Wes +45)? ...] = 036367644171, 
124-215 — 217 = 2 As + 4(a5)? +...] = 020202707317, 
— 412 — 15 4 413 = 2[4q + 44) 4+... ] = (012422519998, 
— 513 + 15 + 217 = Anti + F(z)? +... ] = 008196767203 ; 
from which eliminating 17, we have 


512 + 215 — 613 = ‘092937995659, 
[2+ 315 — 513 = ‘028399474520, 
—412— 15+ 4/13 = 012422519998 ; 


eliminating 73, — 2/2 -- 15 = #223143551312, 
— 16/2- 715 = ‘175710498070 ; 
whence [2 = ‘693147180557, 


15 = 1'609437912426 ; 
ne 110 = 2:302585093003. 
Had the above operation been extended sufficiently far, we 
should have found (272) 
M, _ 1 = 043429 44819 03251 82765 
which, introduced into the last transformation, gives, for the com- 
mon logarithm, 
log. (p+ 2) = Alog.(p + 1) + log.(p — 2) — 2log.(p — 1) 


+ 8685889638 [(——5] 4a (- a nt: ao sy) +a]. (279) 


Employing this form and imitating the above operation for the 
Mopicriay logarithms, 12, 13, i, 17, the student will findt 


* Franceur, Mathématiques pures. 
* The number of digits employed in the calculation should exceed, according to 
the nature of the operation, those which it is intended to retain in the results. 


152 EXERCISES. 


log. 1=0, [10° = 1.] 
log. 2=030102 99957, 
log. 3=047712 12547, 


log. 4= 060205 99913, [log. 4 = 2log. 2} 
log. 5=0°69897 00043, 
log. 6= 077815 12504, flog. 6= log. 2+ log. 3.} 
log. 7=0‘84509 80400, 
log. 8=0°'90308 99870, flog. 8 = Slog, 2.] 
log. 9= 095424 25094, flog. 9 = log. 3.] 
log. 10 = 1:00000 00000 ; (10! = 10.] 
.. log. 11 = 1104139 26852, [p=9.] 
since | 
log. 11 = 2log. 10+ log. 7— 2log. 8-+ ‘86... [3414 4+-4(34y)? 4+. J 5 
log. 12 = 107918 12460, [log. 12= log. 3-+-log. 4.] 
log, 13 = 1411394 33523, fp =11.) 


log, 14= log. 2+ log. 7, log. 15=log. 3+log. 5, log. 16=2log. 4, 
log. 17 = 123044 89214, [p = what ?] log. 18=? 

log. 19 = 1°27875 36010, log. 20=? log. 21 =? log. 22=? 

log. 23 = 136172 78360, log. 24=? leg. 256=? log. =? 

log. 27 =? log. 28=? 

log. 29 = 1:46239 79979, log. 30 =? 

log. 31 = 1:4913616938, log. 32=? log. 33 =? log. 34=7? 

log. 35 =? log. 36=2 log. 37=2 log. 38 =? log. 39=? 

log. 40 =? log. 50 =? log. 60=2 log. 100 =? log. 1000 =? 
log. 10000 =? log. +4 =? log. +¢=? log. 001 =? 


As p increases, the series (278) increases in convergency very 
rapidly ; so much so, that, when pis no greater than 102, the sec- 
ond term will have its first significant figure in the 18th decimal 
place ; and, confining ourselves to 7 decimals, the last.two only will 
have to be obtained by division. 

If we confine ourselves to seven digits, after the computation has 
been made up to 1000, the remaining logarithms may be readily 
ealculated from the table itself. 

Taking the differences of five consecutive POT as those 
of 100, 101, 102, 103, 104, and the differences of these differences 
or the second differences, and the third differences, we find 


INTERPOLATION, 153 


Ist Dif.(-L) | 2d Dif. (—) | 3d Dif. (—) 


20000000 
20043214 
20086002 
20128372 
20170333 


We observe here that the third differences, 8, 9, are nearly equal. 
We are naturally led to the following problem: 


PROPOSITION IV. 


It is required to determine what function y, is of x, when y, 
depends upon x in such way that, if we attribute to x the particu- 
lar values x =0, 1, 2, 3, 4, the third differences of the corre- 
sponding values of the function, y,=Yo. Yiy Yo Ya: Ya Shall be 
equal to each other. 


If in the first power of z, 


we make ta), 1,2, 'o, 4, 

the first differences, 1, 1, 1, 1, are the same ; 
if wad, 1, 273, 4, 

z* gives z* = 0, 1, 4, 9, 16, 

the first differences are, h, 3, 0, a 

and the second differences 2, 2, 2, are the same; 

if z= 0, 1, 2, 3, 4, 

x® gives x? =0, 1, 8, 27, 64, 

the first differences are, 127 Al9, 37s 

the second differences are, 6, 12, 18, 

and the third differences, 6, 6, are constant. 


Hence we assume 
=A+ Br+Cr?+Drz', 


Attributing to z the particular values 0, 1, 2, 3, indicating the 
corresponding particular values of y, by yo, Ys Yo Ys, taking the 
differences as above, and denoting the first difference, y; — y, of 
the first differences by D,, the first difference of the second differ- 
ences by D,, the first of the third differences by D,, (which, for 
the sake of distinction, may be called the first, second, and third 
differences,) we have, 


154 INTERPOLATION, 


YHA, 
D,=y,— y= B+ C+D, 
D,=C 24+ D-6, 

y=A+B+C+D, 

Y—-Y=B4t+C-34 D7; D;,= D+6; 
Y=A+B-24+C-44+D-8, Ce24+D.12; 

B+C.54+D-19; 

yz= A+ Be34C +e 94D « 27; 
». D=tD;, C=4(D,—D,;), B=D,—4D,+4D,, and A= ; 
hence y, = y+(D, —$D,4+ 4D3;)z +4(D.—D;)z? +4D; «2°, (279) 
or y,=Y%+2[D,—4D,+4D,+ 2 +4(D,—D,+2+4D;)], 


= D, D; D; 
Ob. der ther te helt se Bake dy bee? 


is the function sought. 

For any particular series of numbers, it will be advantageous to 
calculate beforehand the coéfficients of the several powers of z, 
and to attribute to the terms their proper signs. Thus, for loga- 
rithms, the second and third differences, D,, D;, being minus, we 
have 

v= +2 | Di +4D.—2D,— 25+ 2 3), 
or 9¥,=%+2[C,—2z(C,4+2C,)], (280) 
putting C,=D,+4D,—4D;, C,=3(D.—D;), C;=2D;. 
Let us make an application by requiring the logarithm of 100345. 
We have 


nee (3 D, = 0043214, |.°. C, = 0043424, 
~ 940000000, | D2 = (0000426, | C= «0000209, 
Yo = 200 > | D, = 0000008 ; C, = 00000014 ; 


log.100:345 = y, = 2+ 345] 0043424 — +345(-0000209-+ ‘345 
* 00000014)]. 


INTERPOLATION. 155 


Operation. 
14 43424 
‘345 72 
345 43352 
115 345 
‘460 13005|6* 
2095 1734)1 
216|7 
20946 eee 
345 0014956 
20000000 
62°7* pe) Se 
84 20014956 = log.100°345 . 
10 
721 


* 
What is the logarithm of 101°79062 Ans. 2007709. 
Required, the logarithms of 100‘1, 1002, 100‘3, 1004, 1005, 
1006, 100°7, 100‘8, 100°9, 
When it is required, as in this example, to interpolate an arith- 
metical progression of tenths, form (280) may be advantageously 
modified, as follows. Suppose that we have found the logarithm 


corresponding to z= ~ and wish the logarithm for x paid! , we have 

10 10 

(280) 
r 7 re 
Her sedge igi 1007" 1006 
” r+l r?+2rt) ret 3r?+ 3r+1. 
Yar = Yor 10 C; <= Om © is ipsatdciomas (17) SERN 
1 2r+1 3r?+3r+1 
—Y,= C e—_—— « —__— —. « ——__________. , 
Year T¥e = Gi 9 ETO “GGT TG 1000 
C 4G, C; 2C, 3C; 


of dret=I-T 19 Too 1000 "Lioo t+") * jooo J (5!) 


To adapt this form to the computation of the tenths from 100 to 
101, we have 


* Shortened multiplication. Rule? 


156 INTERPOLATION, 


C, = 43424, | .-. 3 = = 4342:4, 
C, 6 
Cre 209, an ae 2:09, 
e C; — 6 ° 
C,= 1; Sapo 001 ; 
5 Yeons = = Ye + 4340:309 — 7r[4+18 + (r ch 1) + 003]. 
Making r = 0, we ire log. 100¢1 = site = log. 100 + ‘0004340 
= 20004340 ; 
r=3, 13008 
i], 4340‘309 gives 4340 
gives 4186 12 
Me OO TO Wee oo oe ean 
4336‘123] gives 4340 |100°4) 2:0017336 
are ae 8 16 
ah Lia SP EM ee SON ae OOS 0) En ame Ne een am 


«- log.100°2 = 2:0008676 ;|log.100‘3=2:0013008 ;/100+5) 2:0021660, 
Adapt (281) to interpolate between 101 and 102, and compute the 
logarithms of 101‘1, 101°2, ..., 101°9. 
For returning from the logarithm to its number, we have (280) 
Yr — Yo 

C,—2(C,4+ 20C;) iaiag 2 
where, it is to be observed, that C, is a very near trial divisor, 
though too great. We may therefore find an approximate value 
of x by dividing by C,, and then, having perfected the aes 
repeat the operation. 


Given the logarithm 2‘0014956 to find the corresponding num- 
ber. 


Operation. 
20014956 344 | 43/3|5/2)14956(‘345 

20000000 C.= 13006 
= 4|3|4|24*)1495/6(+344 115 1950 
1303 1734 
cheek 209°459 eae 
192 344 216 
175 or de 216 
pet 62/8 adie 

17 8/4 
8 ., No, 100:345 


* Shortened division. "2 


CHARACTERISTIC, 157 


If we had divided simply by D, = 43214, we should have found 
‘346, which is near the truth, and the approximation will be still 
nearer as we ascend above 100; so that ordinarily it will be suffi- 
cient to diminish the given logarithm by the tabular logarithm next 
below it, and divide this difference by the difference of the tabular 
logarithms above and below. And when greater accuracy is re- 
quired, the third difference may generally be neglected, whereby 
(280) and (282) will be reduced to 


Y2=Yt+2(D,+4tD,—2 4D) ea (283) 


Further it will be sufficient to correct the divisor by the first 
digit of the quotient, or the nearest to it, which may be found by 
inspection. The operation above becomes 


D,=43214 


43 4 2 7/14956(‘34494 
6 4)13009 


4|3|3|6|3| 1947 
1734 


213 
173 


40 
39 


— 


1 


The Characteristic, or integral part of the logarithm, is not usu- 
ally inserted in the tables, since it can readily be determined from 
the relation 

IW'=2; 
which gives 
for y=0, 1, 2% * By ay 1, Ry. 8h on5 
z=1,,10, 100, 1000, ..., °*l;, ‘Ol, ‘OOL, ...5 

Hence the characteristic is always indicated by the dis- (284) 
tance of the first significant figure from the place of units; + if to 
the left, — if to the right. 

Thus the characteristic, or integral part of the logarithm answer- 


158 EXERCISES. 


ing to the number 365 is 2; because 3 is two places distant from 
the units. Therefore, bearing (267) in mind, we have 

log. 365 = 256229, [See tables]. 

log. 365 = log. 42% = log. 365—log. 10=2'56229 —1, =1'56229, 

log. 3°65 = log. 264 = (2 — 2)56229 = 056229, 

log. 365 = log. +288, = (23 )'56229 = 156229, 

log. ‘0365 = log. 73$55 = (2 — 4)'56229 = 256229; d&c., &e. 

It will be seen from these examples that the logarithm of a num- 

_ ber in part or wholly decimal, is to be found in the same way as if 
it were integral, except the characteristic which is determined by 
the rule above and may be either plus or minus, while the tabular 
part of the logarithm is always plus. 


EXERCISES, ~ 
1°, Multiplication (264). 
Operation.| Multiply by 
Multiply 465 266745 | 366 157, 
by 377. 1‘57634 | 90109 ‘035, 
- ‘007 034. 
Ans. 17530. 424379 
2°. Division (267). 
Operation. | Divide by 
Divide 054 Berg099 | « SHS 4095, 
by 1°75. 024304 | ‘0005 ‘00789. 
Ans. ‘030857. 248935 


3°, Involution (265). 
Operation. | Square 139°75. 


Cube 17'356. 123945 (10399)! =? 113 foal. 
3 353 
Ans, 522852, 3°71835 


4°, Evolution (266). 
Operation. | Find ./365, 2/001, 


Find °/2 : 5/ 3010300 : 
¥ nfo 148702 SeamED (75:00005)', V°, 
ns. Is . —- —— pine ot & 
2, 02, /:002. 


Note, An operation should be so conducted as to restrict the 


EXERCISES, 159 
minus sign to the characteristic. Thus, in finding the square root 
of ‘2, we have to divide 1‘3010300 by 2, which is readily performed 
by observing that 1 = 2+ 1, we have 

13010300 : 2=1'6505150 ; 
so 33010300 : 2=(4 4. 1-3010300) > 2 = 26505150. 


5°, Rule of Three. Find a fourth proportional to the 
Numbers, Operation. 


13°79, 1‘13956 
99:367, 1:99724 
720°25. 2'85748 1‘:005 : ‘00356 : : 79099 : 2 


2057106 3657 5892167 


ee eee eee 
ee 


Ans. 51899, 3°71516 | 3379071 * 7112 ** 1000009 * ° 
6°. Exponential Equations—Compound Interest. 
If a denote the amount of one dollar for one year at compound 
interest, then 
aa = a’ will = the amount of $1 for 2 years, 
a’?a = a® = amount of $1 for 3 years, 
a’a = a* = amount of $1 for 4 years, ..., 
a‘ = amount of $1 for ¢ years; 
.. Pa'=amount of $P for ¢ years = A, is the equation for com- 
pound interest. Taking the logarithms of both sides, we have 
log. (Pa') = log. A, 
or log. P + log. (a‘) = log. A, 
or log. P+ t log. a= log. A, 
the logarithmic equation for compound interest, It furnishes, also, 
an example of the solution of an Exponential Equation, since the 
exponent ¢ may be the unknown quantity, and we have 
; — 108: A—log, P 
log. a : 
Application. In how long time will any sum of money double at 
compound interest, at 7 per centum ? 


PROPOSITION V. 


It is required to develope an exponential function in terms of 
the exponent ; 
y=’, (285) 


160 EXPONENTIAL SERIES. 


in terms of z. Attributing to y and x the increments & and A, we 
have 
ytk=at*=a* eat, 
k=a*(a*—1)=a7{(1+56)*—1], [putting a=1+5}, 


or k= o[1+2b-+Rh—1) > +..—1]3 


The 5. ae b be ; 
F =a [b+ (= 1) 6 B+ (b= 1) (h-2) «et ] 


k , D* . Decale 
VY y=fe = l= | le ae oy a Piened wll, —, +. ]=Aa’*, (286) 


: eo* 5 ge 
putting OM gl are 


2 : 3 a2 4 
2 (gee en ee ee 
[See (268), (269).], That is, the derivative of an exponential is 
found by multiplying the function itself by a constant quantity 
which is the reciprocal of the modulus of a system of logarithms 
to the same base, 
Taking the derivative of (286), or the second derivative of (285), 
we find 
alana \otaAae Ait x Ata? s0ql el At a, plizoitin’y a4 
and making x =0, we have 
Yrz—9— GC = I; Yo=9 = A, Yr_9 = al Ys 0 = A®, ooo 9 
whence we are assured, precisely as in the demonstration of the 
Binomial Theorem, that the function y = a’, is expansible in terms 
affected by integral additive power of z and by no others, so that 
the form 
y= a = (+ Cr+ Caz? 4+ Cyr? +...4 Cha* t+ ons 
is possible, and no other. Therefore taking the derivatives, we 
obtain j 
y = Aa? = Ci +14 Che 22+ Cy + 32? + C, o 4z' +... 
—C, <n +, 
y= ARF ='g Ze Lt’ g+3eQe4+C, 046 Bz2+... 
+C,«¢n(n—1)2"*+..., 
yf = Aa? = Ce 23+ 2014 C,+403-2r+... 
+ C, + n(n —1) (n—2)z"*-+.,., &e., &e., &., 
y=Ara = C, « n(n — 1) (n— 2)... 0 32 Bel 
+ C,...° (n+1)n(n—1)... «3 Qat+... 5 


EXPONENTIAL THEOREM. 161 


and, making z = 0, there results 
A? AE 


C,=1, C.=A, a ae | Carrara» 
AP 
wer Gi cL ee 


which substituted above gives 


Tue EXxpoNnENTIAL THEOREM. 


Az? Ara A's 
yoo al4+de+Tigt ig gt tigg gt) C8 


or y=1+ Ly+5 (Ly) RE each 

This last may be data the Antilogarithmic Series. 

In (287) making M=1, A also =1 and a becomes the base of 
the Napierian system, which it is customary to denote by e, Con- 
forming (288) to these conditions, we have 


a Eo 
in which, making z = 1, there results 
Napierian 3 1 1 es 
as =e=141 + ot part = 27182818. (290) 


Again, as A and z are both arbitrary, we may put Az =1, or x 


i whence (288) becomes 


BS 1 

peg bee Pa ele ee 
rane a_la_ Ila 
Le, Jlog.e le. 1 
is the relation between the constant A and the base a of any 
system. 


+ e4=4, OF A= 


Ls fp (291) 


SECTION THIRD. 


General Laws relating to the development of Functions depending 
on a single Variable. 


PROPOSITION I. 


To find the ratio of the increment of any continuous function 
to that of its variable. 


The Theory, or Arithmetic of Functions, like the rules of alge- 
bra, constitutes a distinct science ; for itis capable of development in- 
dependently of any particular or denominate quantity, being equally 
applicable to problems pertaining to numbers, lines, forces, &c. 
But, as an infinite variety of curves may be drawn upon a plane, 
differing from each other in their laws of curvature in every possi- 
ble way, any continuous function, y, depending on a single variable, 
x, may be represented by the ordinate of a continuous (fig. 48.) 
curve PP,, that is, by a curve that bends everywhere by insensible 
degrees, of which the variable x will be the abscissa; and there 
will be an advantage in such a representation, as the understanding 
will be aided by the geometrical figure. 

We have (234) 


Ri Gi y 
eran KARATE 
ti Hl YAN BN PCA, ee A, 
ep va) a XX X AE, WOOK, BAX, X,)? 


from which it appears that z diminishes continuously with h and 
vanishes at the same time}; since it is obvious that, as P, approach- 
es to coincidence with P by a continuous curve, X, must also ap- 
proximate continuously to a coincidence with X,, when 

ems, Av AVA). ye 

~ (X,X)) + (XX) (XX)? 

Therefore observing that (234,) 


a |- [ the derivative of y a 
EX Yiy= se teat 4 fonction of x, 


we have = eo <1 + SY an +3 i. eC, (292) 


z becomes 


Op 


POLYNOMIAL FUNCTIONS, 163 


The ratio of the increment of any continuous function to that of 
its variable, differs from the derivative of the same variable, by a 
quantity which diminishes continuously with the increment of the 
variable, so as to vanish at the same instant, | 


PROPOSITION II. 


To find the derivative of a polynomial, the terms of which are 
continuous functions of the same variable. 


Lety=u+v-...+ constant, be the polynomial, where y, u, 
v, ..., are functions of 2, viz., 
y = fx, u=f,t,0=f2T, 5 
and let the corresponding increments be, t, 7, ..., h; we have 


ytk=(u+i)+(v+j)+...4 constant ; 


k=itj+.n., 
fe a8 yep 
and RA 
or (292) Yyaset 2 = (Waupet 2) (0 r—pet 22) om 
and, 2, 2;, Zo, »» becoming = 9 when h= 0, 
we have Y yafo= Uuapet Dome tors (293) 


or fixr=(fittfrtfrt+..+ constant)’ 
=ffzrtfortfsr+.w3 i ey 

The derivative of a polynomial, consisting of continuous functions 

of the same variable, may be found by forming the algebraical 

sum of the derivatives of its several terms. 

Illustration. The student has already had particular examples 
of this proposition, as in (247), (248), where 2’, 2’, x’, ... are func- 
tions of x. 

Cor. If the functions f,, fi, fa, ... be all the same and a in num- 
ber, (293) reduces to 

(af, x + constant)’ = af,'z ; 


m 
hence, from fiz= ads Ly 
or from nf, 2 = mfoL, 
we have nf c= mf, zr, 


P (=r x) =fir=™ fis; ie (204) 


ee 


164 INTERMEDIATE FUNCTIONS, 


The derivative of a multiple or submultiple function, is equal te 
the same multiple or submultiple of its derivative. 
Illustration. The derivative ty z”™ is mz”, of Az” isA ema”, 


= 


PROPOSITION III. 


To find a derivative by the aid of intermediate functions. 
When y is a function of u, and uw a function of z, the corre- 
sponding increments being &£, 7 and h, (292) gives 


Sega 

for y= fils S = Vy nyu 2 
Rt 

for U = fy Ty = Wusoe t 223 


k z 
ee Se SY yn yy) * (Wants) + Yyate) © 2 


+ (0's. — for) © Zt 2223 
Yyape= (Y'y = pu) © (u ea tia) > 
making h, and, consequently, kh, z,, 2, = 0. 


So from yf=!f uy U SFr V'= I, 2, 
we find yA, = Cie fu) (ue fo) . (V'9—s52) 3 
and generally Yn SY, le Sh * Deas, (295) 


When several variables are successively functions of each other, 
the continued product of their derivatives, taken in the same order 
of succession, will be the derivative of the first, regarded as a 
function of the last. 

Illustration. See Proposition II., Section First, y =f, z = Az”, 


1 

2=fyr=2". 

Cor. The derivatives of converse functions are recipro- (296) 
cals of each other,* 

For, let y and x be functions of each other, or 

y =fx and r=9¢y, 

then we have (Yyare) * (Le gy) = yay = 1, 

1 
Y y=se 

Illustration. In (285) let x and y change places, and compare 
(286) with (268). 


t 


ao OR ) WEEE ee 
* The chord and arc ofa circle are conversely functions of each other. 


FACTORS, 165 


PROPOSITION IV. 


To find the derivative of a product, the factors of which are 
continuous functions of the same variable, 

Given y=fr; y= ut, u=/f,2,0=f,2; to find y',_,. 
We have yptk=(u+2) (+7) =y+vi4-u+y, 

. k=vityt+a; 


but =Wyopet 2% Ort = (uz, + 2)h, 


and = Dyan, tf Za OF J = (Vz + Za)h 5 


k=v(u, + 2)h+ u(v, + 2)h+ (u, + 2) (ve + 2)h? ; 
oF ofus +2) + ule’ +2) + (ul +i) (oe +h 


and Yq i= [+] = (uv) = vu, + uv,’ 
or (uo) = ed ae) = 
UD fig) B 


In the same way, if » be = st, continuous functions of zr, we 
have 
O yelagy = a FB 
ee ee 


which, substituted above, gives, 


(stu) s' tw 
stu s , t “E wu’ 
(stn) pee a Le 
and generally — “e pope et ‘ (297) 


ois (fitfrf,c..) fit, fat, fex 
fishirfew) fi fe fe 

io Tuma RAE 

omitting the variable z, and employing only the symbols of oper- 

ation; fi» Sos tas ves which may be done, since they are equally ap- 

plicable to any quantity which may be made the independent vari- 

able ; thus, instead of 4/xr « 4/x = \/z, we may write 4/+3/=4/, 


as a general rule. We may enunciate (297) : 


166 POWERS, 


The derivative of a product of continuous functions, divided by 
the product itself, is equal to the sum of the derivatives of the 
functions divided by these functions severally. 


PROPOSITION V. 


To find the derivative of any real power of a continuous func- 
tion. 

If in (297) we make the functions s, ¢, u,..., all the same and 
m in number, we find 


, 


(u") a aye _. n—1 ’ 

“a Fae 7 OF (w) = ne ou. (a) 
If y=u-", u being =f, 2, 
we have qu =u" su =u = I, 


(yur): oy 4 (4) + @ apy 
(297) yu pe ey a ih ok ae 


I ; ru ey! rue eu 
a = ah ee e =—y" 
(2) CA iesoat ats 7 (see 
=—ruT ew, (d) 
If y=fr=u", u being =f,z; 
1 
we have y =v", putting 0= wu", oru=v"; 
.*. (a) or (5), y, = me, -and a= ne", 
eal 1 
or (296), v1, = The non? 
f ' ’ t m 1 ’ 
(295), Y, =Y, 0, * Uy = Mv" yn? Ub, 5 
, ™m a , 
or Y.=—U t,. (c) 


From a comparison of (a), (d), (c), it appears that the same rule 
holds for the derivative of a power of a function, whatever real 
quantity the exponent may be ; and we may write 


[( fz") = n(fay (fe), (298) 
or (Fy) =a fy (FY bes, 
The rule found in the first section for the derivative of any 
power of a variable, holds for the power of a function. 
Indeed (298) embraces (245) ; for if fr = 2, then (fr) =2' = 1, 
and (298) reduces to (2”)' = nz". 


FRACTIONS, 167 


PROPOSITION VI. 


To find the derivative of a fraction, the terms of which are con- 
tinuous functions of the same variable. 


From (297) we have 
(ir Alf A(E +8) fifi fe 


.. putting Si=Su to= (;)- (fa)'s 


[fn (fay = (fa + fa +h (APT 
but (298),  [( fae] == fir fs 
(2) 22 Fraexia fi apfpaah, wpe ~ (299) 


f ek Aah eae en 1, €., 
d 


fe" AF) (fa)? 

The derivative of a fraction whose terms are continuous func; 
tions of the same variable, is equal to the denominator multiplied 
into the derivative of the numerator minus, the numerator multi- 
plied into the derivative of the denominator, divided by the square 
of the denominator. 


PROPOSITION VII. 


In finding the derivative of any continuous function (300) 
y = fx, we may replace the increments k, h, one or both, by such 
quantities, k,, h,, as are separately functions of k and h, and such 
that the final or vanishing ratios k, : k, h, : h, become that of 
unity. 

For, by hypothesis, we have 


fi =1-+ 2,, z, being such as to reduce to zero when & and &, 
become = 0; 
or k,=(1+2,)k, so h, = (1+ z)h 
ky 1+-2, & 1 1+2% 
oni betes! fT (Yy=se+ 2) 3 
1+ 2, I+0 
but when h becomes = 055 nye reduces to 1407 1, and we have 


lel-[el-y-- eB. 


168 FUNCTION EXPANDED. 


PROPOSITION VIII. 


To find the expansion of a continuous function, such that its 
successive derivatives all become finite when the independent va- 
riable reduces to zero. 


Let y= fx be the function. It follows, from a process of reason- 
ing precisely like that employed in the demonstration of the Bino- 
mial Theorem, that no other than integral additive powers of z can 
enter into the expansion; and it only remains (and is sufficient) to 
see if the assumption 

y = Ay-+ Ar + Agr? + Azz? +... , 
is possible, or, what amounts to the same thing, if the coéfficients 
A,, A,, As Ag, ..., are determinable, and, therefore, real. 
Taking the successively derived functions, we find 
y =A, 1+ A, + 27+ A, e 3xr?7-+..., 
y= Age 2®el+tAze3e +A, +40 Bx?-+..., 
y' =Az,e3e2el+A, 2 4¢3¢ Qr+Az-+5 +40 3r7+.,., 
yr =Ape 4e38-2-14+A,+-5264-3 6 QrH..., 
&e., &c., &c., &c., 
Now, if in the above equations we make z= 0, and indicate the 
corresponding finite values of 
Yr Ys Yr Yoo Yi oes 
by Yoo Yor Yor Yo's Yos ane 5 
there results y= A, yi=A,o 1, Yi=Ane Vel, Yi/=A,+3e Qo l,...3 


f ae i x “uj x 
YF Yot tos pM ways prs 
4 
(73 8a aioe i 9h) 
This is essentially Maclaurin’s Theorem, and is very service- 
able in expansions, being more general than the Binomial, which 
it becomes simply by putting y=(a+ 2)”. 


+ yi « 


£2, 


2. 


EXERCISES, - 169 


EXERCISES. 
(log. ry=5 . Qr ooh. +, (Ix?) = — =, [(298), (268), (245)}. 
es 
[(log. 2)*] == SE; .. (Unyty =. 
, (log. 2" =="; (ley = % 
[flee 2)" = mui vy 
, M n—l1 
. [log. (az"+ b)] eee 
. [(ale + by] = ania TFG a Gt 
, (log. a*)' = = »Aa=MA=M. =. =1. — [(286, (296).] 
ey _ (lz) 
» [(lx)"] eee 
» (ufeyy = *G. [(298)] 
2 wy _Mb+es %r+d-3r? +...) 
[Ua+ bz + cx? +...)"] = A as Sa PR [(293)] 
i ne eae i 
log. 2) am (2 24 299 
(los. 7) = (7 -F). Libis 
(tos. el =M. mentite A abimaufat bs 
(3) ]- 2. ([(298) 1] 


a 


GY ]-[¢-F]. 


BOOK SECOND. 


PLANE TRIGONOMETRY. 


SECTION FIRST. 


Trizgonometrical Analysis. 


Construction. Describe the quadrant ABC; drop 1 
the perpendiculars BX, BY, upon the radii OA, OC ; 
produce OB so as to intercept AT and CV, perpen- 
diculars drawn through the extremities of OA, OC, 
in T and V; then: 

Definition 1. The ares AB, BC, are said to be com- wt 
plementary to each other. BC is the complement of Fig. 52. 
AB and AB is the complement of BC; the are 90° — a is the com- 
plement of a and a is the complement of 90°—a; 45°42 and 
45° — x are complementary arcs. 

Def. 2. The perpendicular BX is called the sine of the are AB. 
Hence the sine of an arc is the perpendicular let fall from one 
extremity of the arc upon the diameter passing through, the other 
extremity of the same arc. 

Def. 3. AT is the tangent of AB. 

Def. 4. OT is the secant of AB. 

Def. 5. BY (= OX) is the sine of BC or the cosine of AB. 

Def. 6. CY is the tangent of BC or the cotangent of AB. 

Def. 7. OV is the secant of BC or the cosecant of AB. 

Def. 8, AX is the verst sine of AB, 

Note. The abbreviations of the titles above, either with or with- 
out the period, are employed as symbols of the quantities them- 
selves. ‘Thus, if @ denote any arc less than a quadrant and } any 
arc not greater than 45°, the above definitions give 


TRIGONOMETRICAL ANALYSIS, 171 


sina = cos(90° — a), cosa = sin(90° — a) ; 

sin(90° — a) = cosa, cos(90° — a) = sina ; (302) 
sin(45° + b) = cos(45° — b), cos(45°-++b) = sin(45°—d). 

tana = cot(90° — a), cota = tan(90° — a); &c. (303) 
seca = cosec(90° — a), coseca = sec(90° — a); Kc. (304) 


PROPOSITION I. 


The sum of the squares of the sine and cosine of anarc (305) 
is equal to the square of the radius, or to unity, when the radius 
is taken for the unit of the trigonometrical lines, 

We have 

OB? = BX? + OX? = BX?-+ BY?®, (fig. 52.) 
or sin’a + cos*a = r* = 1, 
when radius 7 = 1. 

Cor. The sine is an increasing and the cosine is a de- (306) 
creasing function of the arc, or the sine BX increases from 0 to r as 
the arc increases from 0 to 90°, while the cosine OX decreases from 
r to O for the same increase of the arc, 

See (194) and observe that the sine BX is half the chord of 
double the arc AB. 


PROPOSITION II. 


The tangent of an arc is to the radius as the sine to the (307) 
cosine ; or, the tangent is equal to the sine divided by the cosine, 
if the radius be taken for unity. 


TA * "Bx tana sina 


We have (aet axes par cou (fig. 52.) 
sina 

or tana = ,rm=l 
cosa 


PROPOSITION III. 


The radius is a mean proportional between the tangent (308) 
and the cotangent of an arc; or, the tangent and cotangent of 
an arc are reciprocals of each other when r= 1. 

For, by similar triangles, we have (fig. 52.) 

AT : AO=CO : CV, or tana : r=r: cota; 
tana cota = r*, = 1, when r = 1. 


172 INCREMENTAL VANISHING ARC, 


PROPOSITION IV. 


The square of the secant is equal to the sum of the (309) 
squares of the radius and the tangent. 

We have ) 

OT? = OA?+ AT®, or sec2a =r? + tan2a, ( fig. 52.) 

The student may obtain other forms when wanted ; as, for in- 
stance, the following : 

The secant is to the tangent as radius to the sine, Also (310) 
secant X cosine = 7? = 1. 


PROPOSITION V. 


An INCREMENTAL VANISHING Arc ts to be regarded as (311) 
a straight line perpendicular to the radius. 


Let AB be the are in question; draw the tan- 
gents AT, BT, intersecting in T, and join OT; 
then will the triangles AOT, BOT, be equal, and 
OT will bisect the chord and arc in P and Q and 
be perpendicular to AB, From the similar trian- 
gles TAP, TOA, we have 

AT _OT_OQ+QT_, , QT, 
AP OA OQ OQ’ 
but QT = OT — OQ = (OA? + AT?)? — OA, which reduces to 


1 

[QT] = (OA? — 0?)*— OA = 0, when the are AQB becomes = 0, 
since then AQ=4AQB=0 and AT = tanAQ=0; therefore the 
ultimate ratio of the vanishing pe [AT], [AP], becomes 


A : 
Ee =14[55 pis that of unity ; 


v4 AT TR) [RAL 
and « | |=[ ae |- 


But (113) the are men is greater than the chord AB, and less 


than the broken line ATB; .°. the quotient sa is greater than 


A ATB ‘ 
AB or unity, and less than AB’ which also becomes = 1, when 
the arc AQB=0; 


DERIVATIVES OF SINE AND COSINE. 173 


.", the ratio ea of the vanishing arc [AQB] to its vanishing 


chord [AB] cannot be less than one, and cannot be greater than 
one, and therefore must be =1; which proves the proposition 
(300), observing that the arc is perpendicular to the radius (177). 

Scholium. It is not stated that the vanishing arc, when employ- 
ed as an increment, merely may be regarded as a straight line per- 
pendicular to the radius, but it is proved that it must be so regarded. 
On the other hand, it is to be observed that we do not affirm that 
the arc will ever actually become a straight line, or that it will not 
always exceed its chord in length, but that, for the purpose pointed 
out, it must be so regarded, in order to reduce it to zero, and there- 
by to eliminate it from the function under investigation. 


PROPOSITION VI. 


To find the derivatives of the sine and cosine regarded as 
functions of the are. 
Denoting the arc by z, the sine by y, and the 
cosine by z, the functions may be represented by 
= LB gly 
which are the same as 


Fig. 54. 


y =sinz, z= cosz, 

Attributing to z the incremental vanishing arc h, (311), and to y, 
z, the corresponding increments, k,—7,(306), the similar triangles, 
whose homologous sides, taken in order, are k, h,—i; z,r=1, y, 


give us 
k Zz 
Vren=[Z]= Foes kes 


The derivative of the sine regarded as a function of (812) 
its arc, is equal to the cosine, radius being unity. 


- ' ie = 
Again we have oe = tay; % Ca; 


The derivative of the cosine regarded as a function of (313) 
its arc, is equal to the sine taken minus. 


Proceeding to the 2d, 3d, 4th, &c., derivatives, observing that 
(294) gives (—f)’ = —(f’) when = = — 1, we find, (312), (313), 


174 DEVELOPMENT OF SINE AND COSINE. 


ae sinz, vu =-+2,=cosr, y; = Ze = 94 Uf = Z, = (—y.)' 
=e i, =—Zyy, =z —z,=+ yy y= 2, =y, 
=e, &e,, &c., &e, 3 1. €., 

Cor. The sine and its derivatives are alternately sine, (314) 
cosine ; sine, cosine; ..., in which the algebraical signs alternate 
in pairs, +, +; —,—; +,+; —,-—; .-.., and the cosine and its 
derivatives are alternately cosine, sine; cosine, sine; ..., in which 
the signs alternate in pairs, also alternating, +, —; —, +;+4+, 


oo eet ae 


PROPOSITION VII. 


To develope the sine and cosine in terms of the are. 


Let y=sin(a+2z) and z=cosin(a+ 2); then are y and z con- 
tinuous functions of the arc z, y= Fr, z= F, x, which it is requir- 
ed to determine. 

In (314) substituting @+-z for x, we find 

y=sin(a+ 2), y' =cos(a+ 2), y’ =—sin(a+ 2), 
y'=—cos(a+z), y*=sin(a+ 2), y’=cos(a+z), ..., 
which become 
Yo = sina, y= cosa, yj’ = — sind, y}' = — cosa, 


yy = sina, yj = cosa, yj = — sina, ..., when r=0; 
2 


.. (301), sin(a + x) = sina + cosa > — sind e — cosa 


1-2 


3 4 5 


zw 


. Hb £ 
REDE ee Eee Gi Ere el 
. ‘ 1 i a4 
Hs HY gimmie acon (315) 


kg © oi 
“7.3. Sa Bee : 5) : 3 ‘ m E 5 ; gt ay ..) +0080(5- 7-5 
x ag’ 
tise Tgs4ik 1. OS38t as Boag! 
In like manner (314) we have 


Z) = COSA, Z =— Sind, Z;’= — cosa, z'= sina, z'” = cosa, ... 


CHANGES OF SINE AND ARC. 175 


id 


“. (301) cos(a-+2) = cosa(1— 7" BY 4 .a.304 


(316) 


76 . x3 7° 
mila Se gts) sine pat ye 5 


lta}, 
Making a= 0, and observing that (306) 
sin(@,_9) = sin0 = 0, and cos(a,_)) =r =1 
there results, (315), (316), 


x 3 7° zi 


MLNS PI ETs PPL ETAT E TS Ly ee I) 
and 

2 6 
Thi 9 agp ei ee eet ee carey 


lie andere whee Qin «6 
and these are the developments required, They were discovered 
by Newton.* 

If’we change z into — z, (317) and (318) pesigini (6 ), (6.) 


sin(— Bin Snag iz 57 a pn =a (ez i. oe 1, oad Ue? nat 


2 
satya eat ws ea = 1-5 ts = oe = COST; 1. G., 


Cor. The sine of an arc changes from + to — as the are (319) 
itself changes from + to —, but the cosine remains still + while 
the arc passes through the value zero, which is in accordance with 
(180). Nothing, however, it is to be observed, has been demon- 
strated in regard to arcs greater than 90°, or z > a quadrant. 


PROPOSITION VIII. 


It is required to express the sine and cosine of the sum and 
difference of two arcs in terms of the sines and cosines of the 
arcs themselves. 


Changing x into — 2, (315) and (316) become 
; } co 2° 
sin(@ —zr) = sina(1— Ts +,-, a _ cosa( 2 — {2903 = oe w) 


x* 
cos(a —z) = cosa(I —“Teat ) + sina( 2 — [e203 =; -) ; 


* Lagrange, Lecons sur le Calcul des Fonctions. 


176 FORMS OF SINE AND COSINE. 


with which and (315), (316), combining (317), (318), there results 
sin(a + xr) = sina cosz + cosa sinz, 
sin(a — ©) = sina cosz — cosa sinz ; (320) 
cos(a-+ x) = cosa cosz — sina sinz, 
cos(a — x) = cosa cosz + sina sing, 
These four forms are constantly recurring in trigonometrical an- 
alysis, and should therefore be committed to memory; they may 
be enunciated as follows: 


I. The sine of the sum of two arcs is equal to the sine of the 
first multiplied into the cosine of the second, plus the cosine of 
the first multiplied into the sine of the second. 

Il. The sine of the difference of two arcs is equal to the sine 
of the first multiplied into the cosine of the second, minus the co- 
sine of the first multiplied into the sine of the second, 

Ill. The cosine of the sum of two arcs is equal to the cosine 
of the first multiplied into the cosine of the second, minus the 
sine of the first multiplied into the sine of the second. 

IV. The cosine of the difference of two arcs is equal to the co- 
sine of the first multiplied into the cosine of the second, plus the 
sine of the first multiplied into the sine of the second. 

Consequences. Making a =z, we have (320) 

Cor. 1. sinQr = 2 sinz cosz, (321) 

Cor. 2. cos%x = cos*z — sin?z ; (322) 
but (305), b c= ics? +} sin? 2 
which, combined with (322) and (321), gives 


Cor. 3, 1+ cos2x = 2cos?z, (323) 
Cor. 4. 1 — cos2z =2 sin*z ; (324) 
Cor. 5. 1+ sin2z = (cosx + sinz)*, (325) 
Cor. 6. 1—sin2r = ao ee — sinz)? ; (326) 
* Cor. To( 1+ sine)? +(1- sine) = 2 cosz, (327) 
Cor. 8. (1-+ sinc)? + (1-— sin2r)* = 2 sing. (328) 


What is the sine of a double arc? of half an arc? the cosine 
of a double arc? of half an arc? Enunciate (323), (324), (325), 
(326), (327), (328). 

Adding and subtracting forms (320), and making a+r=p,a—z 
=g,and .. a=+(p+q), r=4(p — q), we have 

Cor. 9. sinp + sing = 2 sint(p+ q) + cosd(p — q), (329) 

Cor. 10. sinp — sing = 2.cosh(p+ ¢) « sind(p — q), (330) 


—— CU a SS ay See eee” 


SUPPLEMENTARY ARCS. 177 
Cor. 11. cosp + cosg = 2 cos#(p+ q) « cosd(p— q), (331) 
Cor, 12. cosg + cosp = 2 sint(p-+ q) « sint(p — q). (332) 


These forms are useful in the application of logarithms, by con- 
verting sums and differences into products and quotients, 

Dividing (329) by (330) we have (307), (308), 

Cor. 13. Piece es Oa AL gs (333) 

sinp —sing  tand(p— q) 

The sum of the sines of two arcs is to their difference as the 
tangent of half their sum is to the tangent of half their difference. 

By similar processes, other forms, occasionally useful, may be 
developed, as 


cosp-+-cosg_ a) 
Cor, 14. Scag sebae cott(p+q) « cotd(p + q), (334) 
sinp-Esing , 
Cor. 15. ‘siaamashonih tand(p + q), 7 (335) 
Cor. 16, Pee ~ cotd(p+q). (336) 


cosg — cosp 
Making a = 90°, forms (320) become 
sin(90° + x) = cosz, 
sin(90° — x) = cosz ; 
cos(90° + xz) = —sinz, 
cos(90° — rz) = sinz; i. e., 

Cor. 17. The sines of supplementary arcs are equivalent, (337) 
being equal to the cosine of what one exceeds and the other falls 
short of 90°. 

Cor. 18, The cosines of supplementary arcs are numeri- (338) 
cally equal, but have contrary algebraical signs. 

Arcs are supplementary when their sum amounts to 180°, as 
(90° + x) + (90° — x) = 180°. 

In the above forms, making zr = 90°, we have 

sinl180° = sin(90° ++ 90°) = cos90° = 0, 
cos180° = cos(90° ++ 90°) =— sin90° = —1; i.e., 
Cor. 19. The sine of 180° is = 0, and the cosine =—1, (8339) 


Cor. 20. sin(180° + x) = — sinz, (340) 
Cor. 21. cos(180° + zr) = — cosz ; (341) 
Cor. 22. sin370° = — sin90° = — 1, (342) 
Cor. 23. cos370° = — cos90° = 0; (343) 
Cor, 2A. sin(370° + x) =— cosz, (344) 
Cor. 25. cos(370° + xr) =+ sinz. (345) 


12 


178 TANGENT AND COTANGENT, 


Scholium. The consequences evolved by these last forms 
(337) ... (345), are in accordance with the principle enounced in 
(180). Indeed, if we suppose the arc to in- 
crease from 0° to 360°, the sine will pass a -& 
through the value 0 at 180° and again at 360°, 
while the cosine will reduce to zero at 90° and peak: + 
270°; and it is obvious that the same correlation \\~ 
of values will be repeated in a2d, 3d, &c., circum- Fig. 55. 


ference. The algebraical sign of the tangent will be determined 
from the relation (307). 


\ PROPOSITION IX. 


It is required to develope the tangent and cotangent of the sum 
and difference of two arcs in terms of the tangents and cotan- 
gents of the arcs themselves. 


Consulting (307) and (320), we obtain 
sin(a+ 6) | sina cosh + cosa sinb 
ag pee bi — + 5 cosa cosb — sina sind 
sina _ sind 
cosa ' cosb 
sina sind’ 
1— ‘-— 
cosa cosb 
dividing numerator and denominator by cosa cosb ; 
tana + tanb 


tan(a + b) = 7 EAR AB’ (346) 

is one of the relations sought. 
So tan(a — b) = anes (347) 
and (308), cot(a- 5) = eee OO a eee . (348) 
Cor. 1. tan2a = 4 doe (349) 
Cor, 2. cot2a = ee : (350) 


Resolving equation (349) in reference to tan a and consulting 
(309), (310), we find 


SCHOLIUM, 179 


4 

—1+(1+tan?2a)? — — 

Con S. “tats ema Pee a Ree as 
tan2a tan2a 


1 — cosa 


" aasinge 


Cor. 4. Cota = cot2a+ (1+ cot?2a)* 
= cot2a+ cosec2a 
2 
= I+ costa : (352) 


sina 


° 
9 


Making @ = 45°, cos2a will=0, sin2a =1, and we shall have 
(351), (352), 


Cor. 5. tan45° = 1 = cot45°. (353) 
whence, making a = 45° in (346), (347), (348), we find 
1 + tanb 
te) — ° 
Cor. 6. tan(45° + b) = yeaa (354) 
5 . cotb= } 
Cor. 7. cot(45° + b) = ant ine i (355) 
Making 2a = 90° + w in (351), we get 
1+ si 
Cor. 8.  tan(45° + 4u) = eee (356) 


Scholium. Other forms, serviceable in turning sums into pro- 
ducts, and vice versd, may be found; for example, if we make 
p = 90° in (329), we get 

1+ sin. g = 2sin(45° + 49)cos(45° — 49) 
= 2sin(45° + 47¢)sin[ 90° — (45° — 4¢)] 
= 2sin(45° + 4¢)sin(45° + 4¢) 


= Qsin2(45° + 4). (357) 

So 1 — sing = 2cos*(45° + 4q) = 2sin?(45° — 49) ; (358) 

1 + cosp = 2cos? 4p, 1 — cosp = 2sin? 4p. (359) 
Combining (329), (330), (331), (332), and (321), we find 

sin*p — sin*g = cos*p — cos*q = sin(p+q)sin(p—gq), (360) 

also cos*p — sin?q = cos(p + q)cos(p — 9). (361) 


The student will also find 
sin(a+5) cotb+ cota  tana-+tand. 


sin(a—b) cotb— cota tana —tanb’ (362) 
sin(a+b)  _cothtcota — tana+tanb | (363 
cos(a= 6) +1-+cotacoth 1+ tana tanb’ ) 


180 DENOMINATE EQUATIONS, 


cos(a-+ 5) cothb—tana 1 — tana tand_ 


cos(a—b) cothb+tana 1-+ tana tanbd’ (At 
eo Tr ir eR a 
tana ee tan?(45° +- $9) ; (365) 
Gite a) Pe 
lalonaal T cot? tp ; (366) 
1+sing _ sin?(45°-+49) 1 —sing _ sin?(45° —iq) | (367) 
Ll+cosp — cos*4p =” ~1—cosg — sin*3g 0’ 
tana + tanb = sles?) ; cota + coth = See ; (368) 
cosa cos b sina sind 
Bw 
tana + cotb = poate cota + tanb = aE A (369) 
cosa sind sind cosb 
PROPOSITION X. 
All denominate equations are homogeneous. (370) 


A denominate equation is one involving denominate quantities, 
such as length, surface, volume, weight, time, velocity, and the 
like, referred indeed to a unit of measure, but distinguished from 
abstract quantities or mere numbers, 

By clearing of fractions and transposing, it is evident that any 
abstract or numerical equation may be represented by 

abc... [n factors] + dgb.C, ... [n, factors] + a3b;c3... [m3] +... = 0; 
for, if there were any powers they would be embraced in the pro- 
ducts of equal factors, such as ab = aa = a?, abc=aaa=a’...; 
a, * b, = d,* d= a3,..., &c.; and we may suppose the equation 
freed from radicals by involution. Now a, 6, ¢c,..., da b,..., being 
numbers, may represent the quotients of any denominate quantities 
divided by their unit of measure, or we may have 


Es OL ee 
OF Fy ~ MM? roo ie tt, 12> VT’ eee gy 
thus, if A =a line 15 feet in length and the unit of measure be one 
A 15 feet 
"M8 feet 
values in the equation above, we have 
At RS ain Agote. 
M ° M e Mu eee [+35 e ik oe [Nq] +... = 9, 
or ABC .,.[n]+A,.B.C, ... [m] + M™™-+ A,B; .. [ns 
é M3 a i = 0, 


yard =the number 5; and so on, Substituting these 


UNIT OF MEASURE, 181 


clearing of fractions, on the supposition that the equation has been 
arranged so thatn > n, >; >.... The last equation is homoge- 
neous, being of the nth degree, or containing n factors in each term, 
and, as it is denominate, the proposition is demonstrated. 

This theorem may be serviceable to those not yet well practised 
in algebra, by detecting errors. For, if we begin a problem with 
a denominate equation, all the following equations being denomi- 
nate, will be homogeneous, and if any one, as that containing the 
result, want this homogeneity, it is an index of error in the op- 
eration, 

But a more important application is the restoring of a quantity 
which has disappeared from a denominate equation by being as- 
sumed as the unit of measure. Thus, if M = 1, the above equation 
becomes 


ABC «...[n]+ A,BoC, + ... [ng] + A,B,C; +... [ng] +... = 0, 


and the homogeneity disappears ; to restore it, however, it is obvi- 
ously necessary and sufficient to introduce the unit of measure, M, 
as a factor, into each term affected by an exponent which is the de- 
ficiency of the term in degree. 

Suppose we are to restore the radius in (317) ; the first member, 
sinz, is of the first degree, every term of the second must be the 
same, which requires 

a 2 x' 
1.2.3 T.. or 1. 78 
The first equation in (320) becomes 
r sin(a + x) = sina cosz + cosa sinz, 
when the radius is restored. So (346).when radius = 7, is 
r*(tana + tanb) 
r? — tana tand ° 


+35 1e 


sinz=2r— 


tan(a + b) = 


It is recommended to the student, as an exercise, to restore the 
radius in all the preceding forms, and to inspect the geometrical 
equations which have occurred in regard to their homogeneity. 


PROPOSITION XI. 


It is required to develope the arc in terms of its tangent. 


Let the arc AB, as it is the function, be indicated by y and its 
tangent AT, being the independent variable, by x; it is required 


182 COMPUTATION OF T. 


to find the function y = fz, that y is of x. Give 
to x the vanishing increment h, to y the cor- 
responding increment hk, and draw J through T 
perpendicular to TO and terminating in the secant 
drawn through the extremities of k and h. . Simi- 
lar triangles give us ( ita 


[l= eMaees LE oa cee ae 


. T om we" 
4 eee tere 


whence, returning to the function (246) we have 


Fig, 56. 


PO et ae x! 
y=Er ges roe ote shims arb constant; 


but z and y vanish together, 
0=0- constant, .. constant = 0; 


= 2" 
YREo sie cach hei pa maveipe tie yee (371) 


which is the required relation. When radius = I, (271) becomes 
y = tany — 4 tan®y + 4 tan’y —4 tan’y+,—,.... (372) 


PROPOSITION XII. 


To compute the semicircumference Tt (pi), when radius is made 
unity. 

For this purpose Machin puts tany = 4; 

is 

(B49) any =, ye oat 

Ze A 120° 
1—()? 119” 
but (353), tan45° = r=1, 


tan4dy — tai45” 1 
: prot eA Gd TTS F ei hopiesto g 
Saale ehainiga din 1 +-tan4y tan45° 9-239” 


whence, by substitution in (372), we ce 
are y tany=+4 =t- +(4)° a 


tan4dy = tan2(2y) = 


os bees $(z45)? Hoy trues 


br = 45° = dy — (Ay — 45°) = 408 — 4) yy ad 
—[as5— tats)? +, —, we} 


and arc(4y— 45°) tan dy—159) - 


COMPUTATION OF T. 


Operation. 

+ Terms. | 
1°, 4 = «2/+200000000000000 | 
3°. (£)5=(4)3(4)?="008-‘04—+00032) 000064000000000 
5°, ‘000000512/+000000056888889 
7, 0000000008192) 000000000063016 
9°. ‘00000000000131072/000000000000077 

‘200064056951982 

— Terms, 
2°, (4)3 = (4) (4)? = 2 + 04 = 008|002666666666667 
4°, ‘00001 28)‘000001828571429 
6°. ‘00000002048 | ‘000000001861818 
8°, ‘000000000032768 *000000000002185 

10°. ‘0000000000000524288) 000000000000003 


‘002668497102102 


‘197395559849880 
4 


4ft—4(4)2 +, —,...] = +. /*789582239399520 
(2 004184100418410 


dif, 


+ |*0000000244 16592 
— |*000000000000256 


4n= |785398163397446 
mT = 3)‘1415926535897/[ 84] 
or 7 =3)/'141592653589793, 
correcting the last digits by an extension of the work. 

Cor. 4 = 90° = 14570796326794896, 
An = 30° = 0'523598775598299, 
10° = 0/174532925199433, 
1° = 0:017453292519943, 
6’ = 0°1 = 0:001 745329251994. 


— [sts —4(a3s)* +> aie ifs 


PROPOSITION XIII. 


t 


To compute the trigonometrical lines. 


Combining (374) with (317) and (318) we obtain 
(0174533)? 


sinl° = (0174533) — ~~ |: 


pel ed 
NW © Ol = 


—,.. = 01745, 


183 


The order of computation is in- 


dicated by 1°, 2°, 3°, .... 


(373) 


(374) 


184 TRIGONOMETRICAL LINES. 


(0174533)? 


1.3 +,—... = ‘99985, 


and cos1°=1— 


Hence, (315), (316), 
sin2° = sin(1° + 1°) = ‘01745 [i-S ae —~——+,-, | 


+ 99985} (‘01 heres penser .. |= 03490, 
2.3 
cose? = 99995 [1 — =P saat sl 
— 01745 [ (1...) it pee he. —, » |= ‘99939 ; 


sin8° = sin(2° + 1°) = 05234, 
cos3° = cos(2° -+ 1°) = ‘99863 ; 
&c. &c. &e. 

The processes just indicated may be advantageously used, the 
first for the computation of the sine or cosine of a small arc when 
a large number of decimal places is required; the second for in- 
terpolating between sines already calculated and set down in a 
table ; but, as two multiplications in each operation are requisite, 
a better method may be employed in making up a table, where a 
single constant multiplrer will be sufficient. 

In (329), making p= (m+ n)a, ¢ = (m—n)a, we have 

sin(m + n)a + sin(m — n)a = 2 sinma cosna, (375) 
a form that will be occasionally serviceable. If, for instance, we 
make m= 1, n= 1, we get 
sin2a = 2 sina cosa, 
a form already obtained ; m = 2, n= 1, gives 
sinda = 2 sin2a cosa — sina 
= 4 sina cos*a — sina 
= 4 sina(1 — sin’a) — sina 
= 3 sina —4sin’a, 

- The student may put m=3, n=1; m=4,n=1; &e.; m=83, 
n=2; m=4, n=2; m=4, n=3; S&e.; and find the results, 
If we make a = 1, we get 

sin(m + n) + sin(m — 2) = 2 sinm cosn, 
which might have been obtained by adding in (320); making 
n= 1° and putting r = the constant multiplier 2 cos1°, we have 
sin(m-+ 1) =r sinm — sin(m— 1), 


TRIGONOMETRICAL LINES, 185 


a convenient form for computing the sines. Making m= 1°, 2°, 
3°, ..., we have 
sin2° = rsinl1° — 0, 
sin3° = rsin2° — sinl°, 
sin4° = rsin3° — sin2°, 
&e, &c. 

It is recommended to the student to find r correct to 7 or 8 
places, and execute the computation just indicated, employing the 
method of shortened multiplication. 

We proceed to show in what way certain sines may be express- 
ed in finite terms. 


In (327) and (328), making z = 45°, we have 
(i+ 1jz+ (1 — 1)? = 2 cos45°, .. cos45° = Abe = a : 
ee 
and =(1+1)?—(1—1)? =2 sin45°, .. sin45° = eae, Cae 
Ze LR 
In (331) making p = (m+ n)a, g =(m—n)a, we have 


cos(m + n)a-+ cos(m — n)a = 2 cosma cosna, (376) 
cos(1-+ 1)a-+ cos(1 —1)a = 2 cosa cosa, 

or cos2a = 2 cos*a — 1; 
cos(2+ 1l)a=2 cosa cosa — cos(2 — 1)a, 

or cosda = 4 cos*a — 3 cosa, 


If, in the last form, we put a = 30°, there results 
cos90° = 4 cos*30° — 3 cos30°, 
or 0 = 4 cos°30° — 3 cos30°. 
‘ 0 = 4 cos*30° — 3, 


sin60° = cos30° = xe ; 


es cos60° = sin30° = (1 — sin?60°)? =. 
Again, 4 cos*18° —3 cos18° = cos3 + 18° = cosd4° 
= sin36° = 2 sinI8° cos18°, 
4 cos?18° — 3 = 2sin18°, 
ia 4(1 — sin?18°) —3 = 2 sin18°, 
or 4 — 4 sin?18° — 3 = 2 sin18°, 
sin?18° + 4sin18° =1, 


cos72° = sin18° = da . 3 
rt 
4 (1042+ 5t)* 


sin72° = cos18° = (1 — sin*18) 4 


186 TABLE OF SINES. 
In (328), making z = 15°, and observing that the minus sign is 
to be employed when z < 45°, we have 
(1+4)2? —(1 —4)? =2 sin 15°, 
1 a 
cos75° = sinl5° = —— (,/3— 1). 
via nae 
The same form will give us the sine of 9°, then 
sine 3° = sin(18° — 15°) = sin18° cos15° — cos18° sin15°, 


and the cosine of 3° being known, the following table of sines and 
cosines may be calculated (320)* 


; 3?+1 re | 1 
sin 3° = cos 87° = 8. Qi (5 —1)-—3— 6+ 5+)’. 
sin fe ox dive Sines ee Ty eee (6 — 5)? 

8 4. 2 
1 1 1 


sin Se ses Bees ppt olor pSaeb')t 


3 1 
; Ons 0 — (5s — 
sin 12° = cos 78° = 3 © I+ 7H 


sin 15° = cos 75° = a (3* — 1). 


sin 18° = cos 72° = is (5¢ — 1). 


4 pay Zit] 

Oo 0 — + e as fe 
sin 21° = cos 69 5. yO sos Magnan 5! — 5? 
sin DAcwaibonibooe t= (ee an he I « 5 — 5 

8 4.2 i 
| 1 tae 
sin 27° = cos GSS st eo (5¢ — 1) +4(5 + 54)?, 


sin 30° = cos 60° = d. 


3 cm 
sin 33° = cos 57° = er (5'—1)+ al (5 + Bi), 


: 1 ne 
sin 36° = cos = So oe (5 — 54)”, 
3+1 3] 4 
sin 39° = cos 51° = & oi (5' ++ 1) — a (Ga By 


* Library of Useful Knowledge. 


TABLE OF SINES: 187 
tiv 3? 
sin 42° = cos 48° = — 4(5' — 1) to (5+ 5')’, 
sin 45° = cos 45° = * 
» (6 1) (Bt 


sin 48° = cos 42° = 


=i or sale Bi)’. 


sin 51° = cos 39° = a 


sin 54° = cos 36° = +9 +1). 


sin 67° =Weuaiggo Se Bek (5§—1)+ ms zie : (5-45)? 
3} 

sin 60° = cos 30° = 3 
; 1 4 

sin 63° = cos 27° = ae (5! — ae vy 

sin 66° = cos 24° = 4(5' + 1) + ——._ (6— 5h)?, 


4.2 3 
2 — yi 
sin 69° = cos 21° = Bah (8-41) + on (6 —5#)?. 


Bo 
sin 72° = cos 18° = a (5 +58)". 


sin 75° = cos 15° = - (3#+- 1). 
2 


sin 81° = cos 9° = a (5!-+1)+4(5 — 54)?. 


sin 78° = cos 12° = $(5'— 1) + 


cele, unary a . 1? 
sin 84° = cos 6 Pry shh) cagregy (or 5) i 


sin 87° = cos oe pelea —1y 4st) 


=: (5 +58)". 


sin 90° = cos 0° = 1, 


Other sines may be interpolated by (315) or by the method of 
differences, and the tangents will be found by dividing the sine by 
the cosine. 

It is recommended to the student to execute several of the com- 
putations indicated above, carrying out the work to ten or fifteen 
decimal places, and employing a shortened method of extracting 


188 LOGARITHMIC SINES, &C, 


the square root as well as in multiplying and dividing. It will be 
advantageous to free the denominators of radicals; for example, 
let it be required to find the sine of 75°. 


Operation. 


sin75° = as (3?-+ 1) = co Saale 


2.2 
3)1°7320508076 2 = 14142135624 
1 27320508076 
27)200 28284271248 
189 9899495937 | 
— 424264069 
343)1100 28284271 
ih 707107 
3462) 7100 eae 
6924 4 
cE te. 8 
3/4|6|4/0|5) 1760000 2 
ear 0°9659258511 = sin75°. 
27712 How many digits may be de- 
74 pended on? 
263 
242 
21 
21 


But a table of sines is comparatively of little importance, as the 
logarithms of these numbers are generally preferable in practice. 


PROPOSITION XIV. 


To compute the logarithmic sines and tangents. 

Restricting (317) to the fourth power of xz, we have 
sing = 2(1 — 42? + +4572"), 

.. log, sing = log. r++ log. (1 —t2? + +2") 


= log. ¢ +-M[(—40*4+44:2")'!— H(—42%)"]_ [(270)] 
ot, Seog gir x*(1 zt =) 
= . 7 6 . 30 ’ 


LOGARITHMIC SINEs, &C., 189 


log. sing = log. z — No. to $2‘8596331 + 2log. zr 

+ log.[1 + No. to (2log. 2 — 1:477)]}. (378) 

As an example, let it be required to find the logarithm of the 
sine of 5°, ' 


We have (374) 
xr = 5° = 08726646 ; 


log. + = 29408474 


2log. x = 3:8816948 [ 3882 
28596331 | 1:477 


——EEE 


log. 15000254 = 00000103 | 4'405 
47413382 LNo. ‘000254 


No. 00005512 


log. sin5° = 29402962 

Form (378) should'not be employed when the arc exceeds 5°, 
and the last term, log. [1 + No. to (2log. 2 — 1'447)], may be omit- 
ted if the arc be less than 3°, 

Imitating the process above, we find 

log. cosz = — No. to $1'3367543 ++ 2log, 2 + log.[1 + 
No. to (2log. x — 0778) ]?. (379) 
Operating as in the last example, we obtain 
log. cos5° = — 0:0016558 = 19983442, 


The logarithmic tangent will be found from the relation 


e 
= —_——,, or log, tan = log. sin— log. cos 
cosine ’ 8 8 8 , 


and the logarithmic cotangent results from 
cos ; 
couse » OF log. cot = log. cos — log. sin. 


Thus log. sin5° = 29402962, 
log. cos5° = 1‘9983442 ; 
fs log. tan5° = 2+9419520, 
and log. cotS° = 140580480. 
In order to avoid minus characteristics, 10 is usually added; thus 
in most tables we find log. sind° = 8*9402960. 


190 ARC FUNCTION OF SINE. 


Dividing the first of (320) by sina, there results 


atl = cosz + cosa sinz = cosz(1 + cota tanz), 
.. log. sin(a+ 2) = log, sina + log, cosz + M(cota tanz 
—tcot?a tan’z +, —,...). (380) 


By a similar process we find 
log. cos(a + x) = log. cosa + log, cosr — M(tana tanz 

+ 4tan’a tan’z+,—,...). (381) 
Let a= 5° and x = 0°] = 67, then 

+ log. sina=2‘9402962 -log. M  =1‘6377843 (1) 

+ log. cosr=1‘9999993 | log. cota =1'0580482 (2) 


+ No. =0:0086639 | log. tanr=3‘2418778 ) (3) 

— LNo. =0:0000864 39377103 

+ 14No, =00000011 42376363 2[(1)-+(2)]+(1) 
, log.sin5°1=2'9488741 65375623. 3[(1)+(2)]-++(1) 


_ Calculate the log. cos5° 1, 

In order to avoid the accumulation of errors, the computations 
should be recommenced from new points of departure, for which 
purpose the above table of sines and cosines to every 3° may be 
employed. 


PROPOSITION XV. 


To develope the arc in terms of its sine. 
We have , 
; k 1 1 
in Lz ih cai | ORES oy, 4 y 
or (250), = (1) *# + (2) (I) + (— 2?) 


1 


or PEM Es we a Deemer set into 
4a F301 6:2 b 1 taBia 2 1 

: 2 2 6 BS 

eee Mae 2) Tes 2 143 sas 
5: 2 2 ' 2 


1 8 


RESOLUTION OF EQUATIONS. 19] 


1+1.-2 1 
ysothedeatpaege TOs 
141-2 142.2 1 


a. g§. ee oS (COU CLO 7 
eas 2 2 Gaebae 


ow eek te Ne oe 2 143-3 
Ay Sua Meh a |, 4 
1 9 
i gg ee vi a Me ee) 
where no constant is to be added, since z and y vanish together. 


PROPOSITION XVI. 


To resolve the equation. 


asinz + b cosr = ¢. 


: sinz b 
Putting tanz = =—, (383) 
cosz a 
: sinz c 
we have sing + * coszr = —, 
cOsz a 


* ; *c¢ COSZ 
sinz cosz -+cosz sinz = 


b] 


: ,  CCOSZ 
or sin(c-+2z) = . 


(384) 
(383) makes known z, then (x-+ 2) is determined by (384) and 
finally z. For an example, let 

/216 « sing + ./72 + cosz = 12; we find x = 15°. 


PROPOSITION XVII. 


To resolve the equation. 
sin(z +k) = m sin(x + 1), 
We have (362), 
sin(z + sin(t +k) _ _ sin[x +4(k+ 1) +4(k —1)] 
sin(z+1) ~~ sinfa-+ 4(k + 1) —4(k—1)] 
_ tanfe+4(k+1)]-+ tand(k—1) | 
~ tanfa +4(k+ 1)] — tant(k — 1)’ 


) 
tanf7+4(k + 1)] = Sa , tana! —k), 
l) 


or tan[z +4(k + A = rn: + v) tand(Z » fos 
putting (354) = tanv. 


192 PLANE TRIGONOMETRY. 


The second form will be preferable, when m is such a quantity 
as to be most readily computed by logarithms. ‘What will the 
equations become when k=0? when /=0? The student may 
form an example for himself 


PROPOSITION XVIII. 


To resolve the quadratic equation, 
2 apr = qq °° 
when p and q are such as to require logarithmic tables. 
We have 
c+p==+(p?+q)'=+(p?+p? tan*v)? [putting g= p? tan’v] 
=-+ p(1+ tan?v)§ =+ p seen, 


a 


acljaiediate em 2 
or r+ p Sarak tanv Phas (385,) 
Example. Given 
25'35sin86 ‘25° 2 © 357sin86‘25° 
an 4 Qe a aaa © = See eo a eee 
sinl61‘75° sin7d‘5°sin161°75° 


to find zx. 


SECTION SECOND. 
Resolution of Triangles and Mensuration of Heights and Distances. 


PROPOSITION I. 


The length of a line and its inclination to a second line being 
given, to find its PROJECTION upon that line. : 


Let a be the given line and / 


the line upon which its projec- 2 

tion is to be made; drop the =... JU a 
perpendiculars p, p, from the Gl) rZ iti fe o 
extremities of @ upon; then CEP y*) 2 

a,, the portion of J intercepted Fig. 58. 


between these perpendiculars, is called the projection of a upon l. 
Through the extremity of a nearest /, draw a; parallel to a, and 
terminating in p,, then a,=a; Also produce a to intersect J, 


PLANE TRIGONOMETRY. 193 


making the angle (a,/); from the angular point and on the pro- 
duction of a, measure off the radius, r= 1, and from the extrem- 
ity of r drop the perpendicular intercepting the cosine, cos(a,/) ; 
then 

Gg inl) it COS( db): 1; 
or ligy 30.2%. COS(Gst) 2 4.3 

dy = a cos(d,l), i..€., 

“The projection of a line is found by multiplying the (386) 

line into the cosine of its inclination to the line upon which it ts 
projected. 


PROPOSITION II. 


To find an equation as simple as possible that shall embrace 
the relation existing between the sides and angles of a triangle. 


Let the sides of any triangle be denoted by a, C 
b, c, and the angles respectively opposite by A, ‘ j 
B, C. Then, dropping a perpendicular from C, 
we have (386) 

dz = acosB, Fig. 59. 
and b, = beosA ; 
s c=a,+ b,= acosB + bcosA, i. e., 

Either side of a triangle is equal to the sum of the pro- (387) 
ducts formed by multiplying the two remaining sides into the 
cosines of their respective inclinations to the first mentioned 
line. 

This proposition, obviously little else than a corollary from (386), 
may be regarded as the fundamental theorem in the resolution of 
triangles ; since it gives at once the equations, 

a=beosC'-+ ccosB, 
b = acosC + ccosA, } (387,) 
c=acosB + bcos ; 
from which, by elimination, all possible relations among the sides 
and angles may be drawn. 

If one of the angles, as A, become greater 
than 90°, the corresponding side, b,= bcosA, will 
be minus (338) .. c=a,—b, = acosB — bcos, 
and the theorem still holds good. 


13 


194 PLANE TRIGONOMETRY. 


In order to find an equation embracing but the single angle A 
and the sides a, 6, c, eliminating cos C between the first and sec- 
ond of (387,), and multiplying the third by c, we have 

a* —b* =accosB—becosA, 


and c* —accosB+ becosA; 
424 c? —a?+b*=2bccosA, 
or 6?+¢?—a*+2becosA; 1. e. 


° 


PROPOSITION III. 


The sum of the squares of any two sides of a triangle (388) 
ts equal to the squares of the third side increased by the double pro- 
duct of those two sides multiplied into the cosine of the angle which 
they include. 


a? +b* —c*+2abcosC 


a? +¢*—b*+2accosB (388, ) 
; b? +c —a*4+2becosA J 
Cor.* a* +6%+c? =2abcosC+ 2accosB+ 2becosA. (389) 


PROPOSITION IV. 


To transform (388) so as to be convenient for the logarithmic com- 
putation of A. 
Combining (388) and (324) we have 
2.44 MQyne 
ead. =cosA=1—2 sin?4A, 
2be 


a®—(b?+c?) a*—(b—c)? 
Pic rte gets Si i sae lc 


2be 2be 
_(a+b—c) (a—b+c). 
‘i 2bc 
_ [(at+b—e) (ate—b)]4 
a a a | 20 6 ae 
a eye (390) 
putting hA=4(a+b-+c). 
So costA= irae ‘. (391) 


* Ixxpress in words. 


PLANE TRIGONOMETRY. 195 


and .*, tan, $A = eo. (392) 


a) 

Also, taking the double product of (390) and (391), we have 
(321), ; 
2[h(h — a) (h— b)(h—c)}* 


sinA = he ; (393) 
sind _ 2[h(h—a) (hb) (h—c) 
“fet abc 
sinB  2[h(h—}) (h—a) (h—c)}* 
so SE 
b bac 
Snes = SRS ora tod. 3,2,9i0.A ‘s sinB ; i. ¢. 


PROPOSITION V. 


The sides of a triangle are to each other as the sines (394) 
of the opposite angles. 
If the angle C = 90°, or B be the complement of A, then ~ 
a sinA sind 


—— = = -—— = tan A, or 
6b snB  cosA ? 


Cor. a =5b tanA, (395) 
Again (394) gives (40,3°) 
a+b: a—6::snA-+sinB : sinA —sinB, 
or (333) a+b: a—b:: tan}(A+ B) : tant(A—B); ie. 


PROPOSITION VI. 


The sum of any two sides of a triangle is to their dif- (396) 
ference, as the tangent of the half sum of the angles opposite 
to the tangent of half their difference. 

The abuve theorems are adequate to the solution of all problems 
in Plane Trigonometry ; and these problems may be reduced to 
one or other of the four following cases : 

Case. I. The angles and one side of a triangle being given, to 
find the remaining parts. 
Rule. As the sine of the angle opposite the given side, 
Is to the sine of the angle opposite the required side ; 
So is the given side ‘ 
To the required side. 


196 PLANE TRIGONOMETRY. 


1°. As an example under this case, let it be requir- 
ed to find the distance, z, of an object rendered 
inaccessible by the intervention of a river. For 
this purpose I measure a base line of 10 chains, and “j 
taking the angles at its extremities, I find them to Fig. 60. 
be, the one 80°, the other 70°, 

We have the angle at the object = 180° — (80° + 70°) = 30°; 


sin30° log. sin30° = 1'69897 [—] 
“sin70° log. sin70°=1':97299 [-+] 
10 log. 10= 100000 [+] 


mv Aals.s, log.caus ~ 127402 
*, @ =18'794 chs. 
Case II. Two sides and an angle opposite one of them being given 
to find the remaining parts, 
Rule. As the side opposite the given angle, 
Is to the side opposite the required angle ; 
So is the sine of the given angle 
To the sine of the required angle. 
2°, To illustrate Case II., in the triangle ABC, let AB= 1356 
chs., BC = 7 chs., and the angle A = 25‘3°; required Z C. 


Operation. 
7 0‘84510 
13°56 113226 
sin25‘*3° 163079 
~. sinC, 191795 


C = 55‘877°, or = 180° — 55‘877 = 124'123°. 
The ambiguity of the angle C will be illustrated by dropping 
the perpendicular BP upon AC, and calculating its length; we 
have 


sin90° 100000 
»gin25*3° 163079 
_ 136. 1:13226 
mink? Taine 076305 

BP = 5'795; 


whence, since BP is less than BC, taking the point C’ in AC and 
on the side of P opposite to C, so that PC’ shall = PC, and join- 


PLANE TRIGONOMETRY. 197 


ing BC’, we have BC’= BC. Therefore the two. triangles, ABC, 
ABC’, are alike compatible with the conditions of the problem ; 
and the angles BC’A, BCA [= BC'C], are supplementary. If we 
had taken A = 445°, we should have found log, sinC = 0413282, 
and, as a consequence, the sine of C greater than radius, which is 
impossible. The cause of impossibility will be manifest by com- 
puting the perpendicular BP, which, being found = 95043, shows 
that BC is too short, when A = 44‘5°, to form a triangle. The 
student should make all the computations here indicated. 
3°, In order to ascertain the altitude, LT, of a Te 
tower which I am prevented from approaching, I 
take a station, A, in the same horizontal plane with 
its foot, L, and observe the angle of elevation, TAL, 
at the top of the tower= 35°; then measuring BAX 
AB = 118 feet directly back from A and in a line CAA 
with L, I find the angle of elevation TBA = 25° | 
What is the altitude of the tower, and what was my distance 
from its foot when at the first station. Ans. LT = 157°74, 
4°, An individual, in order to fix the position of a certain point, 
P, in a harbor, selects a convenient place on shore and measures a 
base line, AB = 19709 rods, and finds the angles, PAB = 100‘13°, 
PBA = 37°.. Required AP, BP. . 
The student will compute AP, BP, and then verify by taking 
AP, or BP, for the known side and calculating AB. 
5°. In order to determine the dis- 
tance between two places, A and B, 
situated on opposite sides of a hill, and 
their relative altitudes, I measure the 
horizontal line AC = 15 rods, and take 
the angles of elevation PAK = 37:9°, 
PCK = 30°, P being a flag on the sum- Fig. 62. 
mit. I then measure the base, BD = 11 rods, and take the angles 
of elevation PBL = 40°, PDM = 31‘14°, also the angle of depres- 
sion BDM = 25°. 
Case III. Two sides and the included angle given, to determine 
the remaining parts. 
Rule. As the sum of the two given sides 
Is to their difference ; 
So is the tangent of the half sum of the opposite angles 
To the tangent of their half difference. 


198 PLANE TRIGONOMETRY, 


6°. A surveyor, wishing to determine the side AB of a field, ren- 
dered incapable of direct measurement by reason of an intervening 
morass, runs the line AC, south 38° west, = '7‘75 chains, then CB, 
south 25‘8° east, 1015 chains. 
Produce AC in K and draw the meridian 
CS; then SCK =88° and BCS = 28'8°, 
2(CAB -++ CBA) = 4BCK = 319°, 


1015-1 7°75 1'25285 
10:15 — 7°%5 038021 
tan31‘90 1°79410 
~ tand(A—B) * 292146 
1(A—B)= 4° 46’ 46” 
and 4(A +B) =31° 54’ 00"; 
i" A = 36° 40’ 46”, 


The angle A being determined the solution will be readily fin- 
ished, and we shall find 


AB, S 1° 19' 14” W, 15246 chs. 


7°. Given the following courses, 
1°. AB, N 46° E, 35 chs. 
2°. BC, N 20° W, 55 chs. 
3°, CD, S 35° W, 45 chs., 
to determine DA. 


[Verify by employing the computed value of DA to: find AB 
already known. | 


Case IV. The sides given, to determine the angles. 

Rule 1. log. sintA=}{log. (A— b)-+log. (h — c)—(log. b+log. c)]. 
2. log. cosgA=t[log. h+log. (h—a)—(log. b+log. c)]. 
3. log. tantA=t[log. (A—b)-+log. (h—c)—log. h—log. (h—a)]. 
4, log. sinA={log. A+-log. (h—a)+log. (A—b }rlog. (A—c)] 

+log. 2—(log. b+-log. c). 
8°. Let it be required to find the angles of a triangular field, the 
sides of which are 3001, 2672, 199 feet. — 


PLANE TRIGONOMETRY. 199 


First Operation. 


a = 30025 | h —b = 115°7%5 206352 f+] 
b = 26775 | h —c = 18450 226600 [+] 
c = 19900 
b = 267°75 2,42773 [—1] 
2)76700|  ¢ = 199-00 229885 ] 
38350 2)1+60294 
log. sintA=  1'80147 
1A = 39° 16’ 40:5” 
A= 78° 33’ 21”. 


Second Operation. 


h = 38350 258377 
h—b=115°% 206352 
a = 30025 2,47748 
c = 19900 2:29885 
2)1‘87096 
log. cosk}B= 198548 
B=60° 55’ 40”, 
a Third Operation. 
h—a= 83°25 1:92038 
h—} = 115° 206352 
h = 38350 2°58377 
h—c = 18450 226600 
2)1+13413 
log. tant$C = 1556706 
C = 40° 30’ 42” 
But B= 60° 55’ 40” 
and A = 78° 33.21; 
aS A+ B+ C=179° 59’ 43” 
diff. from 180° 00’ 00” 
by 0° 00’ 17’ = the sum of errors. 


The third operation for the computation of C is obviously un- 


200 PLANE TRIGONOMETRY. 


necessary, unless we wish to test the accuracy of the work, We 
have employed three methods in order to illustrate the rules ; 
sometimes one and sometimes another will be preferable, according 
to the numbers, 
9°, Required to determine the angles of a quadrilateral field from 
the following data: 
AB=56, BC = 76, CD =87, DA = 43, BD = 67. 
10°. Given two sides of a triangle 367'23, 273 chains, and the 
difference of the opposite angles 15‘7°, to determine the triangle. 
11°. Given the sum of two arcs and the ratio of their sines, to 
determine the arcs. 


Let the arcs be denoted by u, v, their sum by e, and their ratio 
by r; we have 


u+v=e, 
and sinv=r sinu, 
sinv tanw 
we - =r, which put = tanw = ; 
sinw Aare i. 22 
‘ sinu-+sinv 1+r_ 1-+tanw — 
+ sinu—sinv l—r 1—tanw’ 
tand(utv) +r 
qc ee = tan(45° w 
1—r tante 
tan}(u—v) = —— « tante = 2 
x ) ea at. 7 “>  tan(45° + w) : 
which makes known 4(u—vd), 
but H(u+v) =e, 


whence w and v are finally determined. 


The student may make an application of the above in the solu- 
tion of the following problem, taken from Davies’ Legendre : 

12°. ‘From a station, P, there can be seen three objects, A, B, and 
C, whose distances from each other are known, viz., AB=800, 
AC = 600, and BC = 400 yards. There are also measured the hor- 
izontal angles, APC = 33° 45’, BPC = 22° 30’. It is required from 
these data to determine the three distances PA, PC, and PB.’’ 

The angles CAP and CBP will be the w and v of the eleventh. 
The student will make the computations, and devise means to sat- 
isfy himself of the correctness of his results. 


13°. Wishing to ascertain the distance, AP, to an inaccessible 
object, P, also invisible from A, I measure to the right and left the 


QUADRATURE. 201 


equal lines AB= AC = 21°37 chains, and the angles, BAC 
= 113° 12, ABP = 65° 36’, ACP = 89° 5. 

How might the principle of this problem be applied to deter- 
mine the distance of the moon, her zenith distances being observed 
by two astronomers, one at St. Petersburg, and the other at the 
Cape of Good Hope ? 


SECTION THIRD. 
Quadrature of the Cirele, the Ellipse, and Parabola. 


PROPOSITION I. 


To find the area of actrcular sector in terms of its radius 
and arc. 

The sector y is obviously a function of its arc Q 
zx, the radius, r, being a constant quantity. Giv- Jo 
ing to z and y the vanishing increments, A and k, asi 


we have (311), ps mele 3 


[kh] =4r -[h]; Fig. 64. 
rr ys [=] =$7r sors": 
., (246), y =4trx-+ constant, 
but You = 03 


; Gard, les 
The circular sector is measured by half the product of (3897) 
its radius and arc, 
Cor. The area of a circle is equal to the product of the (398) 


radius multiplied into its semicircumference. 
Scholium. The celebrated Problem of the Quadrature of the 
Circle* is evidently reduced to the following proposition: 


PROPOSITION II. 


The diameter of a circle being given, it is required to find the 
circumference. 


* See Montucla, ‘‘ Histoire des Recherches sur la Quadrature du Cercle.” 


202 QUADRATURE. 


If in (3871) we make z =7, there results 
y = +tcircumference = (1—4-+4—4t+5 —s wu)? 5 
(circumference), =4(1—4+4—F+, —, «.) © 27. 
So (ctrcumference),.= 4(1 —4+4—t+5 5 en) © Br 
for any other radius, r, ; 


: (circumference), 2r_ Tr. 
; (circumference), rz 12’ 
and if 2r, = 2, and .*. (378), (circ.),. = 2m, we have 
circumference), 2r : 
icing ee me, =—~=T17 5 1. €., 


Qr 2 


The circumference of any circle bears to its diameter (399) 
a constant ratio. 


(circumference), =T + 2r = 3'1415926535897993 « 2r. 


Cor. 1. The arcs of similar sectors are to each other as (400) 
the radii of the respective circles of which they form like parts, 

For, if uw, w,, denote the arcs of similar sectors, or u 
like parts, as the nth, of the circumferences of which 7 


the radii are r, 7., we have 
NU=T ¢ Br, * 
and NUg = Te aN. 3 


‘ 2 
i r Fig. 65. 


iin ME 

Eliminating the semicircumference from (398) by aid of (399), 
there results, 

Cor. 2. Area of circle = tr?; or, the circle is measured (401) 
by the square of the radius multiplied into 7. 

Cor. 3. Circles and the like parts of circles, as similar (402) 
sectors and segments, are to each other as the squares of their 
proportional lines—such are the radii, diameters, circumferences, 
similar arcs and their chords, sines, tangents, secants, and versed- 
sines. 

For let S-+-s and S,-++s, be similar sectors with 
their equal angles made vertical, s, s,, the similar seg- 
ments cut off from the sectors by the chords c, ¢,, 
T, Tx being the radii, Then S+s, S,-+s,, being 
like parts of their respective circles (?) if n(S-+s) 
represent the first, n(S,-+ s,) will represent the second 
circle ; and we shall have (401) 


QUADRATURE. 203 


a(S -+-s)=tr?, 


and n(S, + s,) = tr,” ; 
nS+s) S+s_ r? S 

hig tN A = -—, and (160) = —; 
n( St S:) S+s, 1,” ( ) S, 

‘ S+s S,+s, Ay Sivoujao® ol Sf 
oe lag 8, Oy aise MAUS coe Shacks 
4 2 _., S#+s _ ntS+s) 

s, 8,” *. S,+% n(S,-+ 2)’ 


which shows that the segments, sectors, and circles are propor- 
tional. Further, denoting the arcs of the sectors by u, u,., and 
., the circumferences by nu, nu,, we have 


Be. te Oe sinw — tanw secu 


NU, UU, . CC, Bing, tau,” sect, 
Nero eT faye (ag) Ue ee BIg 


n(Sits) 8, Te (nr)? (nw)? ue oP sin’ 


Q. E.-D. 
PROPOSITION III. 


An Incremental Vanishing Arc of any continuous (403) 
curve, is to be regarded as a straight line. 

Let z be such an are, c its chord, and t, t,, tan- zat 
gents at its extremities, P, P., that is, coinciding c p. 
in direction with z at P, P., and terminating in Fig. 67. : 
their point of intersection. 

We have (387) 

= tcos(t,c) + t,c0s(t,C) ; 
but it is obvious from the implied condition of continuity, that as 
the arc z decreases to its vanishing state and the points P, P,, ap- 
proach to coincidence, that the tangents t, ¢,, chang- a 26; ta 
ing their directions by insensible degrees, will Me 
‘ ‘ Fig. 68. 

come to form one and the same straight line, and 
the angles, (¢,c), (t,c), decreasing to zero, their cosines become, 
[cos(t,c)]=1, [cos(t,,c)] = 1; 

oe [e] =[4]+ [4]; 

SPH 

and (113), Zz ay : Q. E. D. 

Scholium. Itis obvious that if the curve were ~# 


other than continuous, no such conclusion as the 
above would result. we 


204 QUADRATURE. 


PROPOSITION IV. 


To find the Derivative of the Segmental Area of any contin- 
uous curve referred to rectangular cobrdinates. 


Let Y be segmental area in question, and K 
its increment ; we have (403), (146), | y “ine 
. a 


Fig. 70. 
[K]=(h] Hy y+(a)s - [F ]=9+ 408, but i =05 


y'= [=|- y, which = fz ; i. e., 


The derivative of the segmental area of any continu- (404) 
ous curve is equal to the ordinate of that curve regarded as a 
function of the abscissa. [Y' =y = fx.] 


PROPOSITION V. 


To find the area of the Ellipse. 

Let Y be an elliptical segment embraced by the semi- 
minor axis and the abscissa 2; we have (404), (203), 
(250), ial 


dG) G8) fet ) 


2.3 
eee) ep ah — 1) ip 
° =< ween ig eae ._— — e 
Gee): ¥=0(2 i 3a s" Bal 2.3 
az 
apshat (405) 


the area sought, no constant being added, since Y,_, =0. 
If we make } =a, there results 


ae aly e411) (FQ) 
= —t-. ae 6 toa beak a 
Y. a2 yee 2 5a! 2.3 
x7 
wget? gd np (406) 


the corresponding segment of the circle circumscribing the 
ellipse ; 


Fig. 72. 


QUADRATURE, 205 


b 
a, Fed Xe it 02 Coe aoe a Y,. [Enunciate. ] (407) 
Making z =a and multiplying by 4, we find 


2(z— 1) 


(Ellipse), ,= a(1 BHD pe | 


rn) 


sik eae a any m)aes (409) 


Ellipse : Circumscribing Circle :: 6: a@:: 2b : 2a 
:: Minor Axis : Major Axis. (410) 


(Ellipse),, , = Ma (Circumsc, Cir.) == eta?=abert. (411) 


(Circle), = mb? ; 


. (411), (Liltpse),,.2° (Circle), £: a:; 6. (412) 


PROPOSITION VI. 


To find the area of the Parabola. 
We have (232) 


y’ = y = QUyra® 3 
$+1 
Y=23p7 . it = 2+ Wprrte = tyr; ie, (413) 
2 


The Parabola is two-thirds the circumscribing rect- (413,) 
angle. 

Scholium. It is obvious that the quadrature of the 
circle and ellipse can only be obtained approximately, 
while that of the parabola is exact. 


Fig. 74. 


206 PROXIMATE AREAS. 


PROPOSITION VII. 


To find the proximate area of any continuous curve. 
Suppose the equation of the curve, referred to rectangular co- 
ordinates, to be 
y= Ay+ A,r t+ Anz? + Agxi +... 5 (414) 
then Yo=y=A,+ A,r+ Apr?-+ Azz? +... 
Y=A.r+4A zr? +4A,7?+14A,7' +... + constant, 
but, if the area Y become =0 when x=0, which condition is 
always admissible, since the origin may be taken at pleasure, we 
have 
Y= A.r+4Aj,r?+4A rz? +4A,7!t+... (415) 
But, since three points, P,, P,, P., determine 
with considerable accuracy a curve of moderate 
extent, we will take the foot of the first ordinate, 
Yo, as the origin, and the abscissas, 2, =h, x, = 2h, 
so that the corresponding ordinates, y, Y,, Y2y 
shall be equally distant from each other; .*. mak- 
ing z=0, h, 2h, we have (414) 
Yo= Ay, 
=A,+ A, + h-+ Ay + h’, 
Yo = Ay + A, ¢ 2h-+ Ay » 4h? ; 
“ Yi—Yo=A,+h+ A, ae 
and Yo—-Y= " oh-+ A, 
A —2y, + Yo = Ay ¢ 2h? ; 
but when z= a we have ae 
Y= A, wate 4h? +44, « 8h? = 2h(A, + Ash + 4A, « h*); 
ae = PAY +3 * HY2— 2% + y)], 
or = $h(ty + 24, +44). (416) 


lf we pat ¢ the same method 
of admeasurement and notation, we 
have 


= 3h pe cain. 
Y,=#h (44. + 2y,+ 44), 

Sh (44, + 2y;+ ty), 
&c. &c. 


Y;, — Bh(4Y 202 + 2Y2n—1 + SY on) 3 


EXERCISES, 207 


oe Yop ¥yt Voto + You. = F4(Syo + 21 + Yot 2ya+ ys 
Ht Syst vee + Yonot 2Yan—1 + FY 2n)« (417) 
This beautiful and useful theorem is due to Simpson, The stu- 
dent should enunciate it in common language. 


EXERCISES. 


1°, Required the diameter of a circle having ten linear chains in 
circumference to every square chain in area, 

2°. A square plate of silver, 3 inches on the side, is worth $4. 
What is the value of the greatest circle that can be cut from it? 

3°. Had the plate in 2° been an equilateral triangle, what would 
have been its value? 

4°, The two sides including the right angle of a right angled 
triangle, are three and four rods; what is the area of the circum- 
scribing circle? 

5°, Determine a circle circumscribing an isosceles triangle, the 
two equal sides of which, including an angle of 36°, are 15°15 chs, 
each ? 

6°. The equal sides of an isosceles triangle embrace an angle of 
473°, and the area of the inscribed circle is one acre. Determine 
the triangle. 

7°. A circular plate of brass, 20 inches in diameter, is worth 
$3‘75. What is the value of the three greatest and equal circles 
that can be cut from it? 

8°. Required the area of a circular segment embraced by an arc 
and its chord, the length of which is 5‘87 chs., and the breadth 1:35? 

9°, The dimensions of an elliptical fish-pond are 10 and 15 rods. 
What is its area? 

10°. The ordinate of a parabolic segment is 3 chains, and the 
corresponding abscissa 7 chains. Required the area. 

11°, Required the area of an elliptical segment embraced between 
the semiminor axis and an ordinate = 5‘657 chains, and having a 
breadth of one chain. 

12°. Wishing to ascertain the cross section of a river 100 yards 
from water’s edge to water’s edge, I take soundings every 10 
yards, and find them to be in yards: 

Y= 0, y, = 12, y, = 2041, y; = 253, y, = 284, ys = 29'9, yg = 2959, 
Y, = 261, ys = 209, y, = 128, y,) = 9. 


BOOK THIRD. 


SURVEYING. 


SECTION FIRST. 


Description and Use of Instruments. 


To determine the boundaries of lands, to delineate them in maps, 
and to compute their areas, constitute the Art of Surveying. 

The instruments employed in determining the boundaries of 
lands are, the Chain, for measuring the lengths of lines, the Sur- 
veyor’s Cross to determine right angles, and the Azimuth Com- 
pass, or the Theodolite, for fixing the inclinations of lines to each 
other and to the meridian. 


The Chain. 


Gunter’s chain, as has already been observed, is 4 rods or 66 feet 
in length, and centesimally divided by a hundred links, each, con- 
sequently, equal to 792 inches. 

It is a maxim in land surveying that every instrument, whether 
for measuring lines or angles, must be used in a horizontal po- 
sition; for it is the base, or the projection of the field upon the 
same horizontal plane, that is desired, 

The projections of the bounding lines are usually obtained by 
carrying the chain in a horizontal position, as 
represented in fig. 77, where 1, 2, 3,..., are the 
successive positions of the chain. 

When the inclination of the ground is too 
great to admit of a whole chain, a half or a quarter may be taken, 
and in all cases the proper position of the elevated extremity should 
be determined by a plumb line or by the dropping of a stone. 


CHAIN, 209 


The chief points to be attended to in chaining are, 1°, to keep 
the chain in a horizontal position ; 2°, to avoid straying from the 
line ; 3°, to record without error the number of chains. 

The first condition will be secured by the surveyor supporting, 
when necessary, the middle of the chain, and directing the elevation 
or depression of its extremities. 

The second condition will be readily attained by the foreman 
fixing his attention upon the flag-staff or object of sight, thus draw- 
ing the chain constantly in line. His march will be corrected by 
the hindman, who will cry out “ rieuT !”’ “ LEFT!” as the occasion 
may require. 

Error in record will be guarded against by employing ten iron 
pins which should be turned at the top into a small ring, and tied 
with a piece of red flannel, the better to be seen. The foreman 
takes the ten pins and draws out the chain, the hindman, as it is 
near being stretched, cries ** Down !”’ when the foreman, giving the 
chain a wave to bring its parts in line, pulls it tight and puts down 
a pin. Marching on he repeats the same operation, until, coming 
out empty-handed, he puts his foot upon the extremity of the chain 
to secure it in place, and cries ‘TALLY oNE!’’ and the hindman 
responds ‘‘ tally one!” that the number may be fixed in the mem- 
ory, also recording it in some way, as by a notch in a stick or a 
pebble put in the pocket, if thought necessary. He then, quitting 
the hind end of the chain, marches up to the foreman, who counts 
the pins to assure himself of the reception of the ten; when, stretch- 
ing on, the second tally is executed like the first. A field may be 
surveyed by the chain alone, as illustrated by the subjoined 


Field Notes. 

Contour. AB = 237, BC=4‘67, CD=5‘00, DE=4'98, EA=367, 

Diagonals. BD = 4:83, BE = 5°25 chs, 

Required the angles of the pentagon. 

The question naturally arises: ought we not to measure the in- 
clined plane rather than its horizontal projection, since the surface 
of the former exceeds that of the latter? The answer to this ques- 
tion must be given in the negative, and for two reasons; 1°, a uni- 
formity in surveying different lands is desirable, that they may be 
the more readily compared with each other, and it is obvious that 
this uniformity can be attained only by reducing them to their 
horizontal projections ; 2°, the real value of a field cannot exceed 

14 


210 THE COMPASS, 


that of its horizontal projection, since no more soil will rest on an 
inclined plane than on its horizontal projection, and the same num- 
ber of plants will stand upon the one as on the other. 

For let abcd represent a vertical section of the 
soil of an inclined plane; it is only equal to its hor- 
izontal projection efgh, the vertical depth ad being 
supposed equal to eh; and trees will grow as thick 
together on the inclined as on the horizontal plane. 


The Surveyor’s Cross. 


The Cross has been already described and the method of using 
it in surveying pointed out. [See Book Second, Section First. ] 
The student may now employ the chain and cross to survey a small 
field, and then compute its angles. 


The Compass. 


The Surveyor’s or Azimuth Compass consists of : 
a horizontal circle to which are attached sight-vanes, 
and a magnetic needle delicately balanced on its cen- 
tre, by which the vanes may be directed to any point 
of the horizon so as to determine the inclination of 
lines to the magnetic meridian, and, consequently, to 
each other. The degrees, marked on the limb of the instrument, 
are numbered from the north and south points, N, and S., both 
ways to the east and west, designated by the opposite letters, W. 
and E., for a reason that will presently appear. 

To use the compass, set it firmly upon its staff (better and usu- 
ally a tripod), furnished with a ball and socket joint, capable of 
being loosened or tightened at pleasure, by the aid of which and 
two. spirit levels, placed at right angles to each other on the face 
of the instrument, the limb is to be brought into a horizontal posi- 
tion, When this is effected will be known by the bubbles remain- 
ing in the middle of the levels while the instrument is made to 
revolve on its axis. The needle is now to be let carefully down 
upon its pivot by a screw in the under side. See thatit plays with 
its points just skimming along the graduated edge of the limb, 
Turn the vanes into the required direction by sighting at a staff 
wound with a red flag, and held vertically in line by an assistant 
stationed at a suitable distance. Observe if the needle settles with 


Fig. 79. 


THE VERNIER, 211 


a free motion, describing nearly equal arcs, slowly decreasing, on 
each side of a definite point, and if it finally rests at that point. 
Should there be any doubt as to this, the needle must be agitated, 
either by the attraction of a knife, or by gently tapping the tripod 
with the fingers, and be permitted to settle a second time. To in- 
sure a correct position of the needle is the principal difficulty in 
operating with the compass. 

In order to prevent mistakes, that sight should be turned toward 
the flag staff which will bring the north end of the needle into the 
part of the compass marked N., or with the fleur de luce. The 
bearing will then be read off by the forward end of the needle, 
using the letters it stands between, as in the figure, N. 30° E., if 
the course be northerly, or S. 30° W., if in the opposite direction ; 
both ends, however, are to be observed in estimating the amount 
of the angle. Back sights should be taken at each station in order 
to verify the bearings. 

The Vernier or Nonius isa slip of metal, fitted to slide upon 
the graduated limb of an instrument, and to serve the purpose of 
an extended and impracticable subdivision. 

If x denote the value of a division on the vernier, of which 2 
cover 21 divisions of the instrument, we have 

nmz=n+t1, 


z=l1t- 9 
n 
fen 
n 
17 


from which it appears that the distance, x —1, of the first division 
of the vernier from the first division of the instrument, will be an 
nth part of the unit of graduation—the distance, 2z — 2, of the 
second division of the vernier from the second division of the 
limb will be two nths of the unit of graduation, and so on; so 
that, by sliding the vernier along the limb, we shall be enabled to 
measure spaces to the nth part of the smallest divisions of the 


instrument. Thus, if a scale [eG Ee bb pd 


have a division of 10ths, and [Jo |) [2 [3 [a [5 |e |7 [s [9 [wd 
11 of these be covered by a Fig. 80. 


212 THE THEODOLITE. 


vernier of 10 divisions, we have n= 10, and .. r—1 =); of +4, 
Qr —2=-+4, of yj, ...3 so that, by sliding the vernier along the 
scale to make the points, J, 1; 2, 2; 3, 3, ..., agree in succession, 
there results the measures ‘O01 ; ‘02; ‘03, .... In like manner, if 
30 divisions of the vernier attached to the compass, cover 29 half- 
degree divisions of the instrument, we shall be enabled to measure 
angles to the minute. In some compasses the vernier is attached 
to the extremity of the needle, and, being carried along by it, the 
degrees and parts of a degree are read by simple inspection ; in 
other instruments it is on the outside of the limb, and fastened by 
a clamp-screw from below, which must. be loosened, when the ver- 
nier will be driven by a tangent screw adapted to the purpose. 


The Theodolite. 


The Theodolite consists essentially of a verti- [ Frontispiece. | 
cal and horizontal circle, for the purpose of measuring angles in 
altitude and azimuth. It has, like the compass, spirit levels, by 
the aid of which and screws, the azimuth circle may be brought 
into a horizontal position. When this is accurately accomplished, 
the theodolite is ready to measure any horizontal angle having the 
angular point at its centre, provided all the parts of the instrument 
have been carefully adjusted, To this end, direct both the upper 
and Jower telescopes (the first attached to the vertical circle, the 
second to the axis below the horizontal) to the same mark situated 
at a distance in one of the sides of the angle. Observe the posi- 
tion of the vernier upon the limb of the azimuth circle, reading the 
degrees and parts of a degree by one or more microscopes, fitted 
to this end—unclamp the upper telescope and direct it to a mark 
in the second side of the angle, clamping and finishing the motion 
by aid of the tangent screw. Observe, by the lower telescope, 
whether the azimuth circle has suffered any displacement by the 
motion required in making the second observation; if no such de- 
rangement has happened, the difference of the first and second read- 
ings will be the measure of the angle in question. In order to secure 
greater accuracy, the axis of the azimuth circle may be unclamped, 
the upper telescope brought back to the first mark, carrying the 
azimuth circle along with it—the azimuth again clamped, and the 
angle measured a second time. This operation repeated as often 
as desirable, the whole amount of are passed over divided by the 


THE THEODOLITE. 213 


number of observations, will give the angle required with a corre- 
sponding degree of exactness. Wescarcely need say that the eye- 
glass must be drawn out or pushed in till the cross wires, indicat- 
ing the line of sight—line of collimation, as it is called—shall 
be seen distinctly, and that the object glass is to be moved in like 
manner till the mark becomes well defined. Ina similar way, a 
vertical angle will be measured by the vertical circle, having pre- 
viously brought the telescope to a horizontal position and observed 
the reading, which should then be zero. When this horizontality 
shall be accomplished, will be determined by a level attached to 
the upper telescope, or, if no such level exist, by those already 
mentioned. 

There are several adjustments, either permanently fixed by the in- 
strument maker, or, for the execution of which, he furnishes means 
in screws and parts capable of being detached from each other. 

There are five lines that should be respectively perpendicular to 
each other, viz., the vertical axis, or axis of the azimuth circle, the 
horizontal axis, or axis of the vertical circle, the horizontal line, or 
line of collimation when the vertical circle indicates zero, the ver- 
tical wire, and the horizontal wire—or 


Z (Ay A.) = (H,A,) = (Hy An) = (bs As) = (0,4) 
=(h, Hf) = (0,1) = 90°; 

also the circles should be perpendicular to their ¥ 

axes. The method of testing these adjustments 

consists principally in reversing the lines (73) ; for 

which purpose the telescope and horizontal axis, 

one or both, may be lifted from their ys or supports, the object 


Fig. 81. 


and eyeglasses change places, or the vernier plate carrying the 
vertical circle and the telescope, revolved 180°. Also the 

wires, h, v, which form by their intersection the line of col- ic 
limation, being attached to a ring, may be moved to the 

right or left, elevated or depressed by screws from without. 
If we direct the line H to a distant and well-defined 
mark, and, when the telescope is reversed, find the 
sight upon the same mark, we may be assured that 
the line of collimation, H, is perpendicular to the 
horizontal axis A,. This adjustment perfected, 
the horizontal axis, A,, will be perpendicular to the 
vertical, A,, when, passing the line of collimation, Fig. 81s. 


214 VARIATION OF THE NEEDLE. 


H, through two marks having a considerable angu- 
lar distance, and turning the vernier plate 180°, H 
continues to pass through the same points. 

The vertical wire v will be in a plane perpen- 
dicular to the horizontal axis A,, when, moving the 
vertical circle backward and forward, a distant and 
well-defined mark continues accurately on v. In 
order to know whether the horizontal wire, h, be in its proper 
position, or if the line of collimation, H, when the vertical circle 
indicates zero, be perpendicular to the vertical axis, A,, it is only 
necessary to reverse the telescope, and that the wire h is in a plane 
parallel to the azimuth circle, will be determined by a backward and 
forward motion of the vernier plate. When these adjustments shall 
have been perfected, by often repeating them one after another in a 
different order, whether the levels are parallel to the plane of the azi- 
mauth circle will be known by leveling this circle, making the vertical 
circle indicate zero, if its telescope have-a level attached to it, re- 
volving the vernier plate and seeing if the bubbles continue in the 
middle. Whether the level which may be attached to the telescope 
be perpendicular to the axis A,, will be known by bringing A, over 
two of the leveling screws, and then, by aid of these screws flinging 
A, out of level, or by revolving the telescope in its Ys, if it be capa- 
ble of such a motion. 

If, on any occasion, it be desired to make the vertical circle coin- 
cide with the greatest possible accuracy with a vertical plane, we 
may suspend a plumb line before the telescope and observe when 
the line of collimation traces this line. 

The Variation of the Magnetic Needle may be conveniently de- 
termined with the theodolite by the process of equal altitudes. 

Let the magnetic bearing of the sun before noon at a determinate 
altitude be e°, and at the same altitude after noon f°, and suppose 
x = variation; then will the angular distances of the sun from the 
true meridian be, before noon, e+ 2, afternoon, f= z, but these 
distances are equal, 


f-e 
a" 

The equal altitudes may be taken at any corresponding hours be- 
fore and after noon, and in any season of the year, but the most 
favorable time is about the 21st of June, when the sun, being near 
the summer solstice, will change his declination but a few seconds 


Fs te ee oa A 


LEVELING, 215 


between the observations ; and, if these determinations be made 
several successive days, both before and after the 21st, the errors, 
lying in opposite directions, will tend to balance each other; thus, 
suppose the values of z 


for June 18, 19, 20, 21, 22, 23, 24, 
to be DLiy Lay V33 Ly, V5. Ve. B73 
then GS peakais Mi eta aL oe 4 


The observations may be made several times during the same 
day, but the best hours will be when the sun is in a position nearly 
east or west. 

The Art of Leveling consists in finding the difference of eleva- 
tion between two places: and is not only necessary in the con- 
struction of railroads, aqueducts, and canals, but is useful for laying 
the foundations of edifices, and for other like purposes. The op- 
eration may be readily executed by aid of the theodolite, or, more 
conveniently, with the Leveling Instrument made expressly for the 
purpose, and consisting of a telescope, level, and tripod, being in 
all respects similar to the theodolite, except in not possessing the 
graduated circles. The instrument is to be firmly planted midway 
between two stations, situated at a convenient distance, and its tel- 
escope made to revolve accurately in the horizontal plane, when the 
depressions of the first and sec- 
ond stations are to be noted by 
sighting at a graduated staff held 
vertically on these two points in 
succession. In like manner the 
second station is to be compared 
with a third, the third with a fourth, and so on, to the last. 

For an example, suppose it required to find the elevation of the 
fountain, A, above a dwelling house at B, from the following 
notes, in which the back altitudes are marked + and the forward 


ones —. 


216 PLOTTING. | 


Ans. 10°29 feet elevation. 


SECTION SECOND. 
Plotting. 
PROBLEM I. 


Througn a given point to draw a lineperpendicular to a given 
line. 


Join the given point, P, and any convenient 
point, Q, of the given line ; with the middle, O, of 
the line PQ as a centre, describe the semicircum- 
ference PP’Q, cutting the given line in P’; the 
line drawn through P, P’, will be the perpendicular ag 
required, Why? , 

If the given point be P’ in the given line, set one foot in P’ and 
the other in any convenient point, O, out of the line, deseribe a circle 
and draw the diameter QOP ; PP’ will be the perpendicular required. 

Other methods of drawing perpendiculars may be employed, as in- 
dicated in the figures, but the “* Right-angle”’ is preferable in practice. 


Fig. 83. 


Fig. 833. Fig. 834. 


Fig, 839. 


PLOTTING. 217 


PROBLEM II. 


Through a given point to draw a line parallel to a given line. 

Set one foot of the dividers in the given point, 
P, and with any convenient centre describe the 
circumference PBAX, cutting the given line in 
A and B; take the chord PB and apply it from 
A upon the circumference at X ; PX will be the Fig. 84. 
parallel required. (Why?) Do the same thing with the Right- 
angle and Straightedge. 


PROBLEM Iii. 


From a definite point in a given line to make an angle equal 
to a given angle. : 


Around the given angle, A, with any convenient x B 
radius, AB, describe the arc BC; around the given age 


point, P, with the same radius, describe the are © a2 
QR, Q being a point in the given line; apply the & \ 
chord BC in QR; QPR will be the required an- P Q 
gle. Fig. 85. 


PROBLEM IV. 


To construct a triangle, having given its three sides. 

Draw an indefinite line, KL, in the required posi- Q 
tion, and apply one of the given sides, C, from K to , NI 
L; with the other sides, A, B, as radii, describe 
around the centres, K, L, arcs intersecting in Q; eaana Viana 
KQL will be the triangle required. Fig. 86. 

Scholium I. This problem enables us to plot a field when it is 
surveyed by the chain, that is, when its diagonals are known, either 
by actual measurement or by computation. The student will find 
it a profitable exercise to plot the pentagon given in the preceding 
section under the chain. 

Scholium II. The preceding Graphical Problems give us the 
six following Problems of Construction, whereby any geometri- 
cal problem, solved algebraically, may be executed in a geometri- 
cal way. 


218 PLOTTING, 


PROBLEM V. 
To construct the sum of two lines, 
zr=a+b, 
Lay off 6 upon the production of a. Ch mere ote 
We ae 
Fig. 87 


PROBLEM VI. 


To construct the difference of two lines, 
r=a—b, 
Lay off 6 from one extremity of a towards the other, as P. a2 
from the right hand towards the left in the figure ; the re- Fig. 88. 


mainder will be = z. 


We observe that if 6 exceed a, x will be drawn to the .P _ 

left instead of to the right (180), and that_z will be mi- 

' % Fig. 882. 
nus instead of plus in the equation z = a — 8, 


PROBLEM VII. 


To construct the square root of the sum of the squares of two 
lines, 
x = (a?-+5?)2, 
Make aand 6 the sides of a right angled triangle; z /| 
: Ww ib 
will be the hypothenuse. 4 


a 
Fig. 89. 
PROBLEM VIII. 


To construct the square root of the difference of the squares 


of two lines. 
x = (a? —b*)2, 


On the greater line, a, describe a semicircle, and 
from the extremity of a lay off the chord 5; z will aT 
be the chord joining the extremities of @ and d. 
Fig. 90. 


PROBLEM IX. 


To construct a fourth proportional, 


be 
@:0 30: 2) Or ax = bc, or 2 = —, 
a 


PLOTTING. 219 


Inscribe the chord }-+c in a circle of suitable ra- 
dius, then with a in the dividers and one foot in the 
junction of 4, c, cut the circumference ; the inter- 
section will give the position of a, such that its pro- 
duction, intercepted by the opposite part of the 
circumference, will be z. 

This problem may also be solved as indicated by fig, 23,. 

Scholium. If c be made equal to 8, or if a : b:: 6: a, the 
problem reduces to that of constructing a Third Proportional. 


Fig. 91. 


PROBLEM X. 


To construct a mean proportional. 


1 Nee ape ie sat c, or x = (bc)?. 


In the above, make 6-++c a diameter, and z per- 


pendicular to b+ c; then will a of fig. 91 become 
=z in fig, 92. b 


Scholium. This problem solves also VIIL., since wy 
x = (a? — b*)? =[(a +b) (a—)]?, Fig. 92. 
where xz is a mean proportional between a+b) and a—b; again, 


the eighth may be advantageously employed in executing the pres- 
ent—seeing that 


Q — r\2eas 
x= (be) =[(~$*) -("$)] : 
for b is the sum of bei be 
2 2 
and c the difference of Bene Sp ere 


Scholium. It is to be observed that an equation, in order to be 
capable of construction, must be homogeneous (370) ; and, conse- 
quently, that it must not embrace any unlike quantities, such as 
lines and surfaces. Further, all geometrical quantities are to be 
expressed in lines, as these are the only magnitudes which we can 
measure directly ; thus, a surface S by a square, a®, whose side is 
a; a solid V by a cube, a*, having the side a; a ratio r by two 


. m . 
lines, m, n, r= Teil Such an equation as z*-+ 2azr = b cannot be 


constructed, not being homogeneous. 


220 PLOTTING. 


As an exercise, let it be required to construct the general quad- 
ratic equation 


z? + 2axr = be, 
where a, 5, c, are known lines, 
We shall find v=—at[a?+ dep, 


which may be put under the form 
1 


e=—a%[a(a+=)]"; 


*, 1°, construct “ by Prob. IX, ; 


2°, construct a an by Prob. V. ; 


+ 
3°, construct [=(« +3)] ; by Prob. X. ; 


4°, construct 7 = — as+| a (« + Nils by Probs. V. and VI. 


As a second exercise, let it be required to divide a line into Ex- 
treme and Mean Ratio, i. e., so that the whole line, AB, shall be 
to its greater part, AX, as this greater part, AX, is to the less, XB. 


If the parts be denoted by a+ z and a— @, we shall find 


t= — 20 + fa(at-atatata))*, or = — 2a + [2a Q14)?, 

The student will find it a useful exercise to construct the problems 
which he has already solved, as those at the end of Section Third, 
Book Second. He should also endeavor to combine, as much as 
possible, elegance and simplicity in the arrangement and execution 
of the several parts of a complicated construction. ‘Theorems oth- 
er than those involved in the foregoing problems, as (182); (185), 
may frequently be advantageously employed, 


PROBLEM XI. 


To divide the Gy RIeT epee of a circle into equal parts, as de- 
grees. 

First Method. With the required radius describe 
a circumference, to which the dividers, unaltered, 
will apply six times, since the chord of 60° is = to 
radius. Next bisect the arc of 60° thus found ei- 
ther by intersecting arcs described from its extremi- 
ties and a line drawn to the centre, or by trial ; 


Fig. 93. ° 


PLOTTING. 


221 


in like manner bisect this are of 30°, which must be divided into 


three parts by trial; and, lastly, 
the same method. 


Second Method. Having gra- 
duated a quadrant of large ra- 
dius, as above, transfer the 
chords to a single line. This 
line, engraved upon a ruler, is 


denominated the Line of Chords. 


the arc 


99 


» 


2 


of 5° is to be divided by 


00 
&% 


Chords Nf 


Fig. 94. 


To make use of it, we have only 


to describe a circumference with the chord of 60° as radius, and then 
to this circumference apply the chord of the required arc, also 


taken from the Line of Chords, 


Third Method, Lay off the arc from a scale of equal parts, by 


aid of the following 


Table of Chords. Chord 2x = 2sinz, r=1., 


30° 


0‘347 
0365 0535 


0°/0-00010«174 
1°!10:018|0«192 


0‘518/0'684 
0*700)|0‘861 
2°|10°035/0‘209/0'382/0'5510'71 7/0877 
3°!/0°052/0'226|0'399 0'568)0733)|0+892 
4°|/0:070|0:244/0416 0:585)0749/|0'908 
5°||0087|0'261 |0°433 0601 |0+765)|\0:924 
6°||0°105)0+278)0°450 0:618)0782)/0:939 
7°|'05122|0‘296|0‘467 0'635/0'798 
8°|/0°140)0°313|0°454 0-651 /0+814 
'||0°157|05330 0:501 0°668 0*829 


0+845 


0:954 
0‘970 


0984 


Bley hi 26 
12 Xx Six 


Fig. 95. 
80° | 90° | 
1000] 1*147|1‘286|1‘414 
1015) 1*161) 15299) 1 +426 
1*030) 1‘176|1+312)1°439) 
1:045] 1199) 1'325) 15451 
1*060) 1204/1 +338) 1463 
1075) 1'218)1‘351)1°475 
1089) 1°231|1‘364/1+486 
1'377)1‘498 
1118) 1259) 1‘389)1+509 
16133) 1°272)1+402/1'521 


This method will be found particularly convenient for mechanics 
who have at hand a ruler graduated into inches and 8ths, or, better, 
10ths. Thus, with a radius of 20 inches, the chord of 57° 45’ is 

20(0:954 + 2 » 16) = 19°32 inches. 
Fourth Method. Find how many times the chord of 1° is con- 


tained in radius, We have 


chord 1° = 2sin#°® ; 
log. chord 1° = 03010300 
+ 3:9408419 


= 22418719 ; 


222 PLOTTING, 


rad.[=1} ,, 
- and radius = 57‘2965 chord 1°, 
or = 573 chord 1° nearly. 


Hence a circumference described with a radius of 573 taken 
from a scale of equal parts, will be readily subdivided, the chord 
of a degree being one of these parts. 

Circles, semicircles, and short rulers, made of brass or ivory and 
graduated on their edges, are also employed for laying down an- 
gles. A particular description of these instruments, as well as of 
others, usually contained in a surveyor’s case, such as steel points and 
pens for drawing blank lines and describing circumferences in ink, 
is unnecessary, as their use will be obvious on inspection.* But 
the first method, which requires only a pair of dividers, will be 
found quite as accurate as any, and perhaps as expeditious, We 
scarcely need say that all exact drawings should be made witha 
steel point and afterward inked so far as required. A very conve- 
nient and good pen for describing circumferences may be had by 
thrusting one leg of the dividers through a common pen, having cut 
for the purpose a gash downward in front, just above the clear 
part, and another upward on the back side, a little above the point. 


PROBLEM XII. 


To plot a field, having given the lengths and bearings of its sides. 


As an example, let it be required to plot the first of the follow- 
ing Field Notes. 

Draw a vertical line through the middle of the paper for a me- 
ridian, the top of which is to be regarded as north and the bottom 
as south, then the right hand will be marked E., and the left W. 


* A pair of dividers having three legs will be found very 
convenient in copying. See Lerebours’ Catalogue. 


PLOTTING, 223 


Next, with one foot of the dividers 
centrally situated in the meridian 
NS, describe a circle as large as the 
paper will conveniently admit of, 
when the bearings are to be laid off, 
by the preceding problem, upon the 
circumference from the north and 
south points, Na = 365° towards E., 
Nd = 15‘3° towards W., Sc = 46°, 
towards W, Laying the perpendicu- 
lar of the rightangle upon the centre, 
O, and the point of the circumfer- 
ence, a, and applying the straight- Fig. 96. 

edge to its base, transfer the line Oa to a convenient position, a, and, 
having constructed or chosen a scale of equal parts, lay off 
a =15*75 chains from the first station, 1, to the second, 2; this 
done, b is to be drawn parallel to Ob through the second station, 
2, and measured off from the same scale, when, the third station, 
3, being thus determined, the side c is to be laid down in like man- 
ner; and lastly d. 


The Diagonal Scale is con- 

venient for laying down lines, 
also for graduating the circum- 
ference, either by the third or 
- fourth method. The Sector is 
likewise frequently employed, on 10 
account of the ,facility with which 
its unit is varied to suit the dimen- 
sions required in a given plot, sim- 
ply by causing its branches to ap- io 
proach or recede like the parts of Fig. 93. 
a common jointed rule. Thus, if the sector be opened so that from 
10 on one side to 10 on the other shall = 1 inch, then from 20 to 
20 will = 2 inches, from 21 to 21 will = 2‘1, andso on. A sectoral 
scale may be constructed on paper for the occasion, and to any re- 
quired unit, say 10 to the half inch, by first drawing an are witha 
radius = 4 inch, then running off from the centre, with any conve- 
nient opening of the dividers, a line of 10 equal parts and from the 
extremity drawing a secant to the arc, to which the same parts 
may be applied. 


224 PLOTTING, 


A very small pair of dividers going with a screw, or, simply, a 
fork cut in metal by a file, or even in hard wood with a knife, will 
be found convenient. 


PROBLEM XIII 


To reduce or enlarge a plot. 

From any point conveniently situated, draw lines 
passing through the several stations—then draw, in 
succession and terminating in these lines, parallels 
to the sides of the plot. 

The student will exercise himself in plotting all 
the following fields, measuring the last sides and 
their bearings. 


Field Notes. 
I. 


N 365° E | 15°75 chs. 
N 153° W | 20/00 
S 46° W |.30°25 
S 58°24° E | 20°781 


N 80° 45‘53 chs. 
N28 65°23 
N 86° 57°86 
S 32° 50‘00 


S 46° 53' 23” 50‘82 


S 48*4° WwW 
S 785° W | 15 
N 313° W | 2% 
N 35‘6° E | 33 
SnG6OFes 4 E | 23 
S 4° 43'54" E | 23°764 


LAST SIDE, 225 


IV. 


N 69:0° W 
N 284° W 
N 316° 
N 568° 


S 671° 
S 190° 


V. 


In order to determine a line rendered inaccessible by interven- 
ing declivities, I trace the line ABCDEFG through a neighboring 
ravine, and find AB, N 23° E, 25 chs.; BC, N, 31 chs.; CD, N 
5° W, 10 chs.; DE, N 12° W, 15 chs.; EF, N 10° E, 35 chs., to 
the top of the ravine, thence to the second extremity, G, of the 
required line, S 45° W, 51‘87 chs. Required, AG. 


VI. 
Find where the meridian, passing through the middle point of 
the side a of IV., will intersect the opposite part of the perimeter. 
Ans. At a point in e distant 55 links from the extremity of d. 


SECTION THIRD. 
Computation of Areas. 


PROPOSITION I. 


If the sides of any polygon be projected on the same (A418) 
line, the sum of these projections, taken in order with their pro- 
per signs, will obviously be equal to zero. 

Thus if the sides AB, BC, CD, DE, EA, of 
the polygon ABCDEA, be projected on the line 
LL, in ab, bc, cd, de, ea, we evidently have 

ab +be+cd+(—de)+(—ea)= 09. 
And, generally, if we regard the perimeter as 


described by a point revolving about the polygon 
15 


226 LAST SIDE. 


in a constant direction, as ABCDEA, then will the projection of 
this point upon any line given in position, as LL, describe, by its 
motion, the projection of the perimeter, which projection, increas- 
ing and diminishing, will obviously become nothing when the re- 
volving point returns to its first position, as A, 


Scholium. The principle enunciated in (180), applied to this 
problem, serves to distinguish the plus and minus projections, 
which will be found to be measured to the right or left, corre- 
sponding with the motion of the revolving point. Thus, as the 
point revolves through ABC, the projection, measured towards d, 
decreases, becoming = 0, when the point arrives at D, and then 
reappears, measured in the contrary direction, as the point returns 
through DEA. 

It follows that if a, b, c,..., 7, &, J, 
denote the sides of a polygon taken 
in order;:6'y Biy,C 4) p25 9.5 Koad yA MCIr, Pron 
jections north and south, a”, b", c", ...5 
j', k’, Ul’, the corresponding projections 
east and west, and ma meridian line; 
then will the sum of the meridional 
projections, taken with their proper 
signs, be = 9, as also the sum of the 
projections at right angles to the me- 
ridian ; and we shall find 


a+ B+ c++ j +h +1 =0, : 
or acos(a,m) +- bcos(b,m) + ccos(c,m) +-.. 
+ jcos(7,m) + keos(k, 2pm l we (419) 
since a = acos(a,a’) =acos(a,m), b' = cos(b,m), 
l' = lcos(l,m) ; 
and a’ +b" 4+e"4+..497°+k"4 1 =0, 
or asin(a,m) + dsin(b,m) + csin(c,m) +.. 
te paint gem ) + ksin(k,m) + iain( l,m)=0, > (420) 
since a =acos(da,a’) = asin(a,a’) = asin(a,m), 
Ll" = Isin.(1,m) ; 
asin(a,m) + bsin(d,m) +-...-+ ksin(k,m) 
acos(a,m) + bcos(b,m) +... -+ kcos(k,m) (421) 
_ = isin(tm), = tan(J,m) 
~ — Icos(l,m) writ dd 


by which (J,m) becomes known ; 


SIMILAR POLYGONS. 227 


then (420) 7 = a sin(a,m)+-b sin(b,m) +... k sin(k,m) (422) 
— sin(/,m) 

and the last side, 7, is completely determined—that is. its bearing 

and length are drawn from the bearings and lengths of the other 

sides. Enunciate the above forms. How is the denominator of 

(421) formed from the numerator ? 

Cor. 1. If the sides of a polygon, except one, vary insuch (423) 
way as to preserve their mutual ratios and inclinations constant, 
then will this excepted side bear to any one of the others a ratio 
and inclination also constant.* 

For, denoting the constant ratios which 4, c, ..., k, bear to a by 
Ty Tey oo Ty OF putting 

Va? de Cm fap i, de= F,0; 

which is in accordance with the condition that any one of the 
sides, a, b,c, ..., k, shall have a constant ratio to any other, since 
the ratio of 5 to k, for instance, is 6: k=r,a: 7T,A4=7, : Ty we 
have 

asin(a,m) -+ ar,sin(b,m)-+-ar,sin(c,m) +... + ar,sin(k,m) 
acos(a,m) + ar,cos(b,m) + ar,cos(c,m) +... + ancos(k,m) ’ 
Grpiitat( dan jons sin(a,m) + 7sin(b,m) + 7,sin(c,m) 4-...+ 7,8in(k,m) 

; : cos(a,m)-+ 7,cos(b,m)-+7r,cos(c,m)-+...+7%c08(k,m) ’ 
which fraction is obviously a constant quantity, being independent 
of a, or of the absolute length of any of the sides, a, b, c,..., k, 
and depending only upon the constant quantities, 7, 7,5 ...5 Ti 
sin(a,m), sin(b,m), ..., sin(k,m), cos(a,m), cos(b,m), ... , cos(k,m). 

The tan(/,m) being constant, the 7 (/,m) is constant; 

Zs, (l,a), (2,5), (2,c), ... 5 (1,4), are constant ; 
asin(a,m) + ar, sin(b,m) +... + ar, sin(k,m) 
— sin(/,m) 
=(constant)- a. Q. E, D. 

Polygons, which, like the above, have the same number of sides 
proportional in the same order and their homologous angles equal, 
are said to be similar. 

Cor. 2. In similar polygons, like diagonals are to each (424) 
other as the homologous sides ; for the diagonal is obviously in the 
same condition as a last side, J. 

Cor, 3. The perimeters, or their like portions, in similar (425) 
polygons, are to each other as homologous sides or diagonals, 


tan(/,m) = 


* See Variation, Part 1, Book 1. 


228 SIMILAR POLYGONS, 


For,: a=a/b2=70%c=ra, dS 7, s.¢ k= ra,l=r0; 
give a+b+c+d+..tk4+la=(l4tntrotrat-. bret ri). 


Lemma, Similar polygons may be described by the (426) 
revolution of variable radii vectores, preserving a constant ratio to 
each other and the same angular motion. For let the poles be made 
common in O and the radii OP, Op, depart from the same _ [ fig. 99.] 
axis of angular motion, OaA, then will OpP be a straight line for 
corresponding points of the polygons; and we have, since by hy- 
pothesis OP bears a constant ratio to Op and OA, Oa, OB, 
Ob, OC, Oc, ..., are corresponding positions of OP, Op, 

OP : Op=O04A 2 Da=O0B8 1: OF =00 s Ces, 


whence POA, pOa,— POB, p0Ob,— AOB, a0b,— BOC, bOc,—-... , 
are similar triangles, and AP, ap,— AB, ab,— ABC, abc,—..., are 
similar parts of the perimeters of similar polygons. 

Cor. 4. In similar polygons, the perimeters, or their (427) 
like portions, are to each other as the corresponding radii vectores. 

Cor. 5. Similar curves, regarded as defined by a descrip- (428) 
tion altogether analogous to (426), or their like portions, are to 
each other as their corresponding radii vectores. For (427), de- 
pending only upon the condition that the radii vectores maintain a 
constant ratio, is independent of the number and magnitude of the 
sides of the polygons, which therefore may be made to coincide 
with similar curves by less than any assignable quantities, 

On this principle is constructed the Pantograph, useful for 
copying maps, or any kind of plane figures, whatever may be their 
outlines. It consists of a parallelogram 
of rulers, ABCp, jointed at the angles 
and having two of its sides, BA, BC, 
sufficiently produced to admit of pins 
being inserted in them at P and O, in 
a straight line with the intervening cor- Fig. 1012. 
ner, p. If O be fixed, the points, P, p, will describe similar 
curves; so that an exact copy will be obtained, either diminished 
or enlarged, according as the pencil is placed at p or P—and the 
relative magnitudes of the figures may be varied by changing the 
pins, A, C. If p be fixed, P and O will describe similar figures, 
which may bear any ratio to each other, not excepting that of 


equality. 
Scholium I. There will be no difficulty in determining (429) 


EXERCISES, 229 


the direction of the last side, 7, if we observe that, on arriving at 
its first extremity, our progress will have been farther north than 
south, or our LatirupE north, when the sum of the northings 
shall exceed that of the southings, or when 

a +6 -+...+k' or acos(a,m)+...+h cos(k,m)> 0, i. e., +; 
and v, v.; also that our position will be found to the east of the first 
station, or our Departure east, when the sum of the eastings 
shall exceed that of the westings, or when 

a’ +-b"4+-...+k" or asin(a,m) +... +h sin(k,m)> 0, and ». v. 

Scholium I. A part of the tabular labor may be saved by (430) 
making the projections upon and perpendicular to one of the sides ; 
thus, if we assume a as a false meridian, or make (a,m) =O, 
(421) becomes 

6 sin(b,a) +... + & sin(k,a) 
a-+b cos(b,a) +... +k cos(k,a) 
_ bsin(b,a)-++ ... +4 sin(é,a) 
‘- > Sinll,a)x oy dons 

Scholium ITT, By eliminating between (420) and (419) any (431) 
two sides, as k and 7, may be determined when all the other quan- 
tities are given; but it will be better to operate as if 1 were a me- 
ridian, when (l,m) will=0, (a,m) = (a,l), (b,m) = (0,1), (c,m) = 
(c,l), and from (420), (419), there will result 

a sin(a,l)+5 sin(b,l) +c sin(c,l) +... +7 sin(7,) 
vasa Dain(kyl) 
—l=a cos(a,l) + 6b cos(d,l) +c cos(c,l)+...+% cos(k,/). 


tan(l,a) = 


and (422), l 


9 


EXERCISES. 


1°. Required the last side in I. 


Operation. 
15°75 sin 365° — 20 sin 15*3° — 30°25 sin 46° 
15°75 cos 365° + 20 cos 153° — 30°25 cos 46° 
[N, +] [N, +] (s, —]. 


= tan(d,m). 


230 EXERCISES. 


logs. nos’ 1‘19728 130103 148073 
logs. sins 1*77439 142140 1,85693 
logs. coss 1:90518 198433 184177 


log. a’ = 097167. log. b” = 0'72243 log. c’ = 133766 ~ 


» log. a = 1510246 log. b' = 1'28536 log. ci = 1532250 
a’ =a sin(a,m)=-+ 93685 a =a cos(a,m) =+ 12661 


b = 5 sin(b,m) = — 5775 b’= b cos(b,m) = + 19*291 

ce’ = c sin(c,m) = — 21°7600 c’ = c cos(c,m) = — 21014 

numerator = — 17‘6690 denominator = -+ 10938 

. eastings < westings. .. northings > southings. 
log. 17669 = 1:24721 log. 17669 = 1°24721 
log. 10°938 = 103894 log. sin 58:24° = 1:92955 
log. tan (d,m) = 020827 log. d= 131766 
-, (d,m) is S. 58*24° E. d= 2081 


If, in accordance with the second scholium, we assume a for the 
meridian, whose bearing is N 365° E, the NW and SE bearings 
will be increased by 36‘5°, and the NE and SW diminished by 
the same angle; hence Field Notes, No. I., will be transformed 
into 


a N, 000° E, 15°75 chs. 
b N, 51‘8° W, 2000 

c SB, 952, 30°25 

d 


The student will calculate d according to this table. 


2°, Calculate e of IT by (421) and (422). 

3°. Calculate f of III twice, by projecting first on 5 then on c, 
and balance the errors by adding together and dividing by 2. 

4°, Calculate g of IV and verify by a different method. 

5°. Find AG in V. 

6°. Calculate the diagonal drawn from the first extremity of a in 
IV to the second extremity of c. 

7°. Solve VI, finding also the length, ?, of the intercepted me- 
ridian. [1 mtersects e: why 2] 


AREAS, 231 

— 115 sin69°—17 sin28*4°--11 sin31‘6°-+15 sin56‘8° 

= 7 = 055, 
— sin67‘1° 

l= 115 cos69° + 17 cos28*4° + 11 cos31‘6° + 15 cos56‘S° 
— 055 cos67‘1° = 36°444. 

8°. Required the length, 7, and second point of intersection with 
the perimeter, of a line running from Z (a,b) of LV, N 25° W. 


k 


PROPOSITION II. 


It is required to find the area of a polygon in terms of the sides 
and the angles which these sides make with each other. 

The easiest way of ascertaining the : 
area of any polygon (a, 0, c, ..., 7, &, L), 
is obviously to divide it into triangles . 
(A,B), (Ay Cy (Ay dy ony (As j)s (Ay Bs 
and then to compute these triangles. 
But the double area of a triangle is had 
at once by taking the product of its base 
and altitude. ‘Therefore from -A, the 
common vertex of all the triangles, and 
upon their bases, 5, c, d, ..., 7, k, pro- 
duced, if necessary, let fall the corresponding perpendiculars, 
Pos Pes Pas +++ > Pj» Px; there results 

2triangle (A,b) = dp,, 
2tr(A,c) = cp,, 
Ztr(A,d) = dp,, 

&e., «&c., 
2tr(Ay) =jpp 
2tr(A,k) = kp, 

But p, is obviously the projection of a upon the line of p,, p, the 
sum of the projections of a and } upon the line of p., p, the sum 
of the projections of a, 6, c, upon the line of p,, ..., p; the sum 
of the projections of a, b, c, ...¢, upon the line of p,, p, the sum of 
the projections of a, b, c, ..., k, upon the line of p, ; 

.". Py = @ COS(a,p,) =a sin(a,b), since Z (a,p,) + (a,b) = 90° ; 
so p,=a@sin(d,c)+5 sin(d,c), 

pa=a sin(a,d) + b sin(b,d) +c sin(c,d), 

&e., &c., &c., &e., 

p; = a sin(a,j) +b sin(b,j) +c sin(c,j) +... +4 sin(iy), 
p, =a sin(a,k) +b sin(b,k) +c sin(c,k) +... +7 sin(j,/) ; 


232° AREAS, 


therefore, substituting above, and adding, there results, 


[ adh st ab sin(a,b) (432) 


+ 2(A,c) +- ac sin(a,c) + be sin(8,c) 
QP=4 + 2A,d) = + ad sin(a,d) 4+ bd sin(b,d) + cd sin(c,d) 
+ &c. +&e, &c, &e., 


| + 2(A,-) | {+ ak sin(a,k) + bk sin(b,k) +... +k sin(7,-) ; 

. The double area of any polygon is equal to the sum of the 

products of its sides, save one, multiplied, two and two, into the 

sines of the angles formed by the sides belonging to the several 
products. 


Cor. 1. The double area of a triangle is equal to the pro- (433) 
duct of two of its sides multiplied into the sine of the angle inclu- 
ded by them. 2tr(a,b,c) = ad sin(a,6). 


Cor, 2, The area of a paralJlelogram is equal to the pro- (434) 
duct of its dimensions multiplied into the sine of the included an- 
gle (433). 

Cor. 3. When two triangles have an angle of the one (435) 
supplementary to an angle of the other, the triangles are to each 
other as the products of the sides about the supplementary angles 
(433). 

Cor. 4, Combining (433) and (393), 
or 2tr(a,b,c) = ab sin(a,b) 


Q[h(h — a) (h—b) (h—c)]*. 
ab 5 


and sin(a,b) = 


there results 
tr(a,b,c) = [h(h — a) (h—b) (h—c)]23 we. (436) 
The area of a triangle is equal to the square root of the contin- 
ued product of the half sum of the three sides and the three 
remainders formed by diminishing this half sum by the sides 


severally. This furnishes a rule very convenient for the applica- 
tion of logarithms. 


Cor. 5, Similar polygons and their like segments and sec- (437) 
tors are to each other as the squares of their homologous lines, 
whether sides, diagonals, or radii vectores, 

For, preserving the notation under (423) and substituting in (432) 
there results 


AREAS. " 2338 


¢ 


| r, sin(a,d) 7 
+r, sin(a,c) + 7,7, sin(d,c) 
ae = J + r,sin(a,d) + ryrz sin(d,d) + r,rq sin(c,d) + a*, (438) 
| + &c., &c., &, | 
| + r,sin(a,k)+ r.r,sin(b,k)-+...+ rr, sin(7,k) | 
Cor. 6. If the polygon is equilateral, that is 
Cae pac Spy Be Sh 
or fp. Slee = 7, > 7, > 1, 
we have 


+ sin(a,d) + sin(d,d) + sin(c,d) a”, (439) 
+ &c., &c., &c., 
wt sin(a,k) + sin(b,k) =... + sin(7,£) 
Cor. 7. If the polygon be regular, that is, have its angles as well 
as its sides equal, putting 
Zz (a,b) =€, 
and therefore, (a,c) =e, (b,c) =e, 
(ajd) = Se, (b,d) = 2e; (cd) = e, 
&c., &c., &ec., 
(a,k) =(n— 1)e, (b,k) =(n — 2)e, 
(k= (ty 'B)e sce (75h) es 
n-+1 denoting the number of sides of the polygon, there results, 
(439), 


sin(a,b) 
+ sin(a,c) + sin(d,c) 
= 


Fig. 1022. 


[ sine 7 
| -+ sine + sine 
2P,,.., =< +sinde + sine + sine “at, 
+ &c., &c., &c., 
| +sin(n—1)e+sin(n —2)e+sin(n—3)e+ ... a 


or 2P,..,=[(n— l1)sine + (n — 2)sin2e 
+ (n —3)sin3e + ... + sin(n — 1)e] « a?. 
Cor. 8. Making n = 2, we find 


. 360° 
P,= +| @- 1)sin | «a? = 45in60° + a? = 32(1a)?, (441) 


0 


for the area of the Equilateral Triangle. 
Cor. 9. Making n = 8, we find 


P, = 3[2sin90° + sin180°] + a? = a?, (442) 
for the area of the Square, as already known. 


234 AREAS. © 


Cor. 10. Making n =4, we find 
P; = $8in36°(1 -++ 6cos36°) « a? 
I 1,4 
(6+3 e 5?) (2 oh — 2-0 52) 
lie 
for the area of the Regular Pentagon. 
Cor. 11. For n= 5, or the Regular Hexagon, 


P,= a? + 3sin60° = a? « oye (444) 


a’, (443) 


Cor. 12, Similar plane figures, whether bounded by (445) 
straight lines or curves, are to each other as the squares of their 
homologous ares, chords and radii vectores, (437,) (426). 

Scholium I. It is obvious, from the course of the de- (446) 
monstration of (432), that the angles must be all estimated in the 
same direction, either from the right round to the left, or from the 
left to the right. And the rules for the algebraical signs of the 
trigonometrical lines, as everywhere else, are here also to be rigor- 
ously observed, 


Thus, sin211° = sin(180° + 31°) =— sin31°. 
For the computation of the area of the first field, we have 
OD ab sin(a,b) [2(A,d) J 
+ ac sin(a,c) + bc sin(d,c) ; [2(A,c)] 
es a 
Z (a,b) = 36'5° + 153° = 51'8°, 
(a,c) = 180° + 365° — 46° = 170°5°, WwW E 


(5,c) = 180° — (15°3° + 46°) = 118*7°. 


Fig. 1023. 
Operation. 
51‘8° | 1189534 | 239365 
a, c | 170*5° | 1*21761 | 1689562 
b, c | 118*7° | 1:94307 | 2*72383 
a | 15°75 | 1:19728 | 24755 [2(A,d)] 
b | 2000 | 130103 | 7864 ae A,c)] 
30°25 | 1°48073 | 529°45 


2P = 855'64 


P = 427,82 sq. chs. 
= 42782 acres. 


AREAS, 235 


Operation for P of II. 


N 
180° b ce 
aN 80° W Ww E 
bN 2° W L, 
cN 86° E , 
dS 32° E Fig. 1024. 


3:463 1495 | 290:502/290:502=2(4,b) 
2804 3513 | - 63°731 
3576 5612 | 377+191/440:922=2(.4.c) 
3:228 3412 — 169177 
b, d| 150°/T698 9700] 3212 3874 | 163075 
c,d) 62°'T945 9349] 3407 2833 | 255437 249335-2( 4,d) 


45+53|1+658 2977/44] [4 980°759 


a 

5b |65'23)1'814 4474). + ay 4 
90379 ss 

c |57‘86/1°762 3784 379 acres = P 

d 


50°00 


a,b) 78°)1990 4044 
a, Cc 166°/1383 6752 
b, c| 88°'T999 7354 
a, d| 228°/T871 0735 


Scholium II, The operation may be materially shortened (447) 
by making a diagonal the last or excepted side, and computing the 
polygon in two distinct parts; also gross errors will be detected, 
and the slight ones, always incident to mathematical tables and 
instrumental admeasurements, corrected, by changing the diagonal. 

Thus, if we imagine a diagonal joining the extremities of a and 
c in II, we shall find 

(a,b,c) = 365‘711 acres, 
(d,e) = 124669 do. ; 
a (a,b,c,d,e) = 490380. 
Again, let the diagonal join the extremities of 5 and d, and there 


results, 
(b,c,d) = 397'852, 
(e,a) = ~ 92629, 
ate (a,b,c,d,e) = 490381 ; 
so (c,d,e) = 345‘131, 


(a,b) = 145:251, 
(a,b,c,d,e) = 490382 ; 


aS 3(a,b,c,d,e) = 1471'143, 
and (a,b,c,d,e) = 490-381. 


236 AREAS, 


Scholium III. When the sides of the polygon are numer- (448) 
ous, it will be expedient to divide it into parts by one or more 
diagonals determined by computation or by actual measurement. 


EXERCISES, 


1°, Calculate the area (a,b,c,d) of I by calculating the parts 
(a,b), (c,d). 
2°, The same by computing (d,c) + (d,a). 
3°, Balance errors by taking the half sum of 1° and 2°, 
4°, Find the area of I by excepting d. 
5°. The same by excepting a. 
6°. The same by excepting 6. 
7°. The same by excepting c. 
8°, Balance errors. 
9°, The area of II may be calculated by several of the following 
methods : 
(a,b,c) + (d,e), (b,c,d) + (e,a), (c,d,e)-+ (a,b), (d,e,a) + (b,c), 
(e,a,b) + (c,d) ; 
(a,b,c,d), (b,c,d,e), (c,d,e,a), (d,e,a,b), (€,a,6,c). 
10°. Employ some of the fifteen methods following for the 
computation of the area of III. 


(a,b,c) + (d,e,f), (b,¢,d) + (e.f@), (c,d,e) + (F145) 5 
(a,b,c,d) + (e,f')y (b,c ae) )s pie d,e.f) + (a,b), 
(d,e,f,) + (5,¢), (e.f,4,6) )» (f,4,6,¢) + (d,e) 5 
(a,b,c,d,e), (5,¢,d, a (c,d ‘ok a), An e,f,a,b), (e,f,a,5,c), (f,a,b,c,d). 
11°, Write out all the methods for the computation of the area 
of IV, and execute a number of them. 


12°, Calculate the area lying on the west of the meridian which 
passes through the middle point of the side a of IV. 
Ans. 44‘812 acres. 
13°. Compute, by the aid of logarithms and (436), the area of a 
triangle the sides of which are 13‘334, 15°75, 16°024 chains. 
14°. Find the area of the pentagon given under ‘“ the Chain.” 


AREAS, 237 


PROPOSITION III. 


To cut off from a polygon a given area, PBCDX, by (449) 
a line, PX, running from a given point, P, in one of the sides, 
as AB, 


Compute the triangles PBC, PCD, PDE.,..., till 
two consecutive areas, PBCD, PBCDE, be found, 
the first less, the second greater than the required 
area, PBCDX; then will it be known on what 
side, DE, the point X must fall. The areas DPE, 
DPX, being known, the one by computation and 
the other by hypothesis, we have the proportion 


DPE DE 

DPX DX 
for calculating DX, when XP will be determined by the process 
for a last side, 

As an example, let it be required to cut off 200 acres from the 
left of II, by a line running from the first station A. By referring 
to the computation above, we find the point X will fall on c, and 
the operation may be put down as follows : 


tr(A,c) 220461 sq. chs. 3'343332 
tr(A,z) 54749 sq. chs. 2°738376 
_¢ 57:86 chs. 1°762378 
~ 2 . 2 = 14'369 chs. 1‘157422 


PROPOSITION IV. 


To lay out a given area in a triangle of a given form. 


Let A, B, C, denote the angles severally opposite the sides 
a,b, c,; then (432), (337), 
2P = ab sin(a,b) = ab sin, 
2P = be sin(b,c) = be sinA, 
2P =ca sin(c,a) = ca sinB ; 


op — sinB sinC he ge (450) 


sinA 


Cor. 1. If the triangle be isosceles, or C = B, whence 


238 AREAS, 


180°=A+B+C=A-+2B, 
B=C=90°—+#A, 
sinB = sinC = sin(90° — 4A) = cosfA, 
sinBsinC costA costA _ costA costA 
sinA sinA ~ sind A cossA 
there results 
2 
P = cotha. (451) 
Cor. 2. If the isosceles triangle P be applied m times about its 
vertex A, to form* the regular polygon, P,,, we have, 
nme 
Eo m@r= re cottA, 
ma? _ 180° 
or Pa = ae eo (452) 
since mA = 360°. 
What will (452) become for m = 3, 4, 5, 6, 7, 8, ... 2 
For illustration, let it be required to lay out ten acres in a trian- 
gle whose angles are, A = 45°, B = 87°, 
We have (450) 


200 sin45° 14. 
sin87° ae | : 

a = 13‘80:chs, 

For the side of an octagon of 15 acres, we find 


8 a. 
150 = a te cot 8 


or 75 = a? cot224° ; 


+ 
> o=| we | = [75 tan224°]? = 31°07 chs. 


PROPOSITION V. 


To lay out a trapezoid of a given area, P, on a z 


base, a, also given, knowing the inclinations of the » ad be 
° ° ce 
oblique sides, xX, y, to a. Fig. 104, 


We have 
ra sin(z,a) + ry sin(z,y) + ay sin(a,y) = 2P, 
ay sin(a,y) + az sin(a,z) + yz sin(y,z) = 2P, 
zx sin(z,z) + za sin(z,a) + rasin(z,a) = 2P ; 


* The student will construct the figure, 


AREAS, 239 


but sin(y,z) = sin(a,y), sin(z,x) = sin(z,a), sin(a,z) = sin(z,a) = 0; 
the second and third equations reduce to 


y(a + z)sin(a,y) = 2P, 


x(a-+ z)sin(z,a) = 2P ; (453) 
eosin{a,a) 
a so (454) 
by which the first equation reduces to 
J ‘ sin(a,y) 2P ; sin(a,7¥) 
aflaety [2 Saleen sa sin(z,a) sin(z,y) ” ee) 


sin(a,y) 2P sin(a,y) 
sin(z,y) ? sin(z,a) " sin(z,y) 
ted, this equation may be reduced as an ordinary quadratic, but 
better by (385,), observing that 


The coéfficients a « , being calcula- 


ik 
q” sin(a,y) 
A tanv eg p=a4e STIGHI NE (456) 
ype J sin(a,y) | 
1= Sin(aa) * sin(ry) > (286) 
and x thus becoming known (454) determines y, when (453) gives 
(a+ z) and, consequently, z. 
Let it be required to cut off 15 acres by a line parallel to 6 of I. 
Making } = a= 20, the operation will be as follows, 


xz N 365° E id.y pbs LISe7° 


a N 153° W (x,y) = 1705° 

y S 46° W (z,a)= 518° 
log. sin(a,y) = 1943072 
log. sin(z,y) = 1‘217609 


log. sin(a,y) — log. sin(z,y) = 0°725463 


log. 2P = 2477121 
log. sin(z,a) = 1‘895343 


log. 2P — log. sin(z,a) = 2581778 


log. g = 3307241 


—_—_— 


log. g+ = 1653620 


240 AREAS, 
" log. a= 1'301030 
log. p = 2026493 
log, tanv = log. gt — log. p = 1,627127 
log. cosv = 1964237 


—- 


log.(z + p) = log. p — log. cosv = 2062256 


z+ 1063 =+115*4 


Ty OA 
_ sin(z,a) ie ice 
“sin(ay) 7 = 4° 
2P 
eis Z = 21595 


It may sometimes be desirable to execute the above problem 
without the aid of tables, or any other angular instrument than the 
cross. For this purpose measure in any conve- z 
nient position a line, p, perpendicular to a, and 
through its extremity a parallel, b, determining the 
trapezoid (a,5), which should differ as little as one 
may judge from (a,c), the area to be laid out. 
The area of the trapezoid (b,z) thus becoming known, it remains 
only to find its breadth, z; and for this purpose we have 


a(b + 2) jt tem: - 


(24 
Fig. 1042. 


trap « (b,z) = a bn ee 
2bp _ 2p trap(d, z) 
° 2 a a & 
aA cdl rh eee ied Taek (457) 
PROPOSITION VI. 
To cut of a given area,XABCDEFY, Y 


from a given field, by a line, YX, 
running in a given direction. 


Having ascertained what.sides YX ©< v 
will intersect,* find AZ parallel to XY, Fig. 105. 


* This may be done, if preferred, by dividing the plotted field ABCDEF into trian- 
gles, taking the measures of their bases and perpendiculars from a scale of equal parts, 
and finding the sum of their areas; which, however, must not be employed any far- 
ther; for when FZ, ZA, have been found by calculation, or by actual measurement 
in the field, the area of ABCDEFZ must be accurately computed. 


= -.' --_ -* —  - 


AREAS. 241 


and compute the area ABCDEFZ; when the area of the trapezoid 
XAZY becoming known its parts will be found by Proposition V. 


PROPOSITION VII. 


Given the length and direction of the line AB, 
intersecting the lines AK, BL, also given in di- 
rection ; to draw a line, KL, in a given direction 
and intersecting AB in I, so as to make, on oppo- 
site sides of AB, the areas IAK, IBL, equal. CC? 

Representing the lines IA, IB, by a-+z2, a—z, Fig. 106. 
and the angles IAK, ILB, IKA, IBL, by e, f, m, n, we have (450) 

(a+ x)? sine sinz 
sinm 


e ae 
sine \2 sinn 
“ (a@+27)(-— =(a—2 
a (Sam) (a (ae 
. ak e L e e 1 
(Ss) (= ) , ( sine se) 
_ \sinf sinm} | = sinm sinn} |. 


: a : 1 e ° 
sinn\* re sine \2 i+ sine sinf \? 
sin sinm sinm sinn 


(a — x)? sinn sinz | 


= 2tr([AK) = 2tr(IBL) = sinf 


1 — tanv 
or (354), z = ———_-. - a = atan(45° — v), 
eat 1 + tanv ( i (458) 
. sine sinf \7 l 
putting tanv = (Sinn sie ° J 


Scholium I. We observe that (458), being well adapted to log- 
arithmic computations, may be advantageously combined with (449) 
for the solution of Propositions V and VI, also in changing the 
direction of a line between two farms. 

Thus, let it be required to cut off a given area, KUVL, by a 
line, LK, running in a given direction. In the line VL take any 
point, B, and by (449) determine the point A in the line of UK, 
such that AUVB may = the required area, KUVL, and apply (458). 
It will obviously facilitate the operation if it be convenient to re- 
duce B to coincidence with V. The same process will change AB, 
the division line of two farms, into a second required position, KL. 

Scholium II. The intelligent student will not find any difficulty 
in applying the propositions now given to the straightening of a 
broken boundary between estates. : 

16 


242 AREAS, 


PROPOSITION VIII. 


Through a point, P, given in position in a giv- 


en polygon, to run aline cutting off a required 
area, S+8S’. 

By calculation or actual measurement, determine Z8\ 
any convenient line passing through the given point Fig. 107. 


and separated by it into the parts a, 6; also compute the included 
area S+S”; it follows that, since S-+S'’ and S+S" are given 
quantities, their difference, S’'—S" becomes known. But we have 


a’? sinmsint _ 2S! and 5? sinn sinz 4 
sin(m-+i) sin(n +7) 
a? sinm sint 5b? sinn sinz ; A . le 
"yaaa 2) Gna e —S a which put=c, 


a? sinm sinz b? sinn sinz 


28" ‘ 


or Se eS Fe a a ee 
sinm cos? + cosm sin? sinn cos? + cosz sinz 
a* sinm ' 6? sinn 
or ACE? Wy ih Seng TIMES Bike. a) 
sinm cote + cosm  sinn cotz-+ cosn 
Ai a* sinm sinn cott + a? sinm cosn 


— b? sinm sinn coti — b? cosm sinn 
=c sinm sinn cot*z + c(sinm cosn + cosm sinn)cott 
+ ¢ cosm cosn ; 


cot?z + pa an) + Aes a S| cotz 


sinm sinn c 
a? cotn—b? cotm 
= —__________——_— cotm cotn, 
c 
i b+a)(b— a . a*cotn 
or, cot?7 + [ cotn + | cotz = ‘Saat 
if it be convenient to make m = 90°. (459) 


Prana tiebname Woke ontinals 


SOLID GEOMETRY, SPHERICAL GEOMETRY, 
AND NAVIGATION. 


wy wre AR ; 


_ ea @ om , vu We me f » i aghe 
pe pebyguat, ty ue ves Po rained 
Sli eae aE | 

i A, a eedidig ot: oreehs oetgrioondie’ et er bers 
ney. hs ery a ee che ge tan girin lt 


Fin er krin 


Grn mete: stress eliieinta Wk EQST ye eee 


| ae: cht 


N eA an wuss 


“Nano, idtemaantte \earisKoun ) ax199, 
oth tt POORLY ti OR Boe 
.* ell jo eee RO F 


petitbicte StS eee ok 


BOOK FIRST. 


SOLID GEOMETRY. 


SECTION FIRST. 


Planes. 


PROPOSITION I. 


Three points not in the same gr arete line determine the (460) 
position of a plane. 


For, let the plane, P, pass through two of the 
points, as A, B; then, revolving upon these points, 
it will become fixed or determined in position, 
when it shall contain the third point, C, 


Fig. 108. 

Cor. 1. Two intersecting lines, as AC, BC, Aa ee (461) 
the position of a plane. 

Cor. 2. A triangle, as ABC, is always in the same plane (462) 
and determines its position. 

Cor. 3. A plane is determined in position by two paral- (463) 
lel lines, as AP, BQ. 

Cor. 4. Two planes cannot intersect each other in more (464) 
lines than one. 

For if A and B be any points common to the planes P and P,, 
it is obvious from the definition of a plane that the straight line 
AB will lie wholly in both planes, and will therefore be a line of 
intersection, Now the planes, P, P,, cannot have a second line of 
intersection, as ACB; since this hypothesis would reduce the 
planes to coincidence (460), the three points, A, C, B, not in the 
same straight line, becoming common to P and P,. 


246 PERPENDICULAR PLANES, 


Cor. 5. The intersections of planes are straight lines. (465) 
Cor. 6. Planes coinciding in three points, not inthe same (466) 
straight line, coincide throughout. 


PROPOSITION II. 


A line drawn through the intersection of two other lines (467) 
and perpendicular to both of them, will be perpendicular to their 
plane. 

For let p be the perpendicular, a, a, equal por- 
tions of the intersecting lines, b, b, the hypothe- 
nuses to 7p, a, -- p, a, which will therefore be equal, 
and let d be any line drawn through the angle 
(a, a) and terminating in the line m-7n joining 
the extremities of a, a, and divided by d into the Fig. 109. 
parts m, n ; it only remains to show that e, joining the extremities 
of p, d, isa Sai We have (137) 

— d?=mn= b?—e?; 

, aging lade D., (133). 

Cot. 1. ieee one side of a right angle be made a fixed axis (468) 
of revolution, the other side, in revolving, will describe a plane. 

For, if the right angle (a, p) revolve about p as an axis, a will 
be found constantly in the plane of (a, a). 

Cor. 2, Of oblique lines drawn from any point in a per- (469) 
pendicular to a plane and terminating in this plane, the more dis- 
tant will be the greater, those equally distant will be equal and ter- 
minate in the circumference of the same circle having the foot of 
the perpendicular for centre. 

Cor, 3. Planes which are perpendicular to the same (470) 
straight line are parallel to each other; and, con- 
versely, if a line be perpendicular to one of two par- 
alle] planes, it will be perpendicular to the other also. 


Fig. 1092. 
Cor. 4. If from any point without a plane, a perpendiec- (471) 


ular be dropped upon the plane, and from the foot of this perpen- 
dicular a second perpendicular be let fall upon any line in the plane 
the line joining the first and last-mentioned points, will be perpen- 
dicular to the line drawn in the plane. 

For if m =n, d and e will both be perpendicular to m + n. 


PARALLEL PLANES. Q47 


Cor. 5. Through the same point, either without or within (472) 
a plane, but a single perpendicular can be drawn, 

For let e be any line intersecting the perpendicular, p; e is ob- 
viously inclined to d and therefore to the plane (a, a). 

Cor. 6. A plane passing through a perpendicular to a (478) 
second plane, is perpendicular to the same. 

For let d revolve to take up a position perpendicular to a; then 
(p, d) being a right angle, the plane (p, @) is said to be perpendic- 
ular to the plane (a, a, d). i 

Cor. 7. The intersection of two planes perpendicular to (474) 
a third, is perpendicular to the same plane. 

Cor. 8. Lines perpendicular to the same plane are paral- (475) 
lel to each other. 

For let p, ps, ps3, .-., be lines perpendicular to 
the plane, P, and a, b,..., the lines joining the 
points in which the perpendiculars intersect the 
plane; it follows from (473), (474), that the planes 
(p, @), (pz, 5), will intersect in p,; whence 7, pr, 
being perpendicular to a, ps, p3, to b, ... 5 Py Pos Pas 
are parallel to each other. 

Cor. 9. A line parallel to a perpendicular to a plane is (476) 
itself a perpendicular to the same plane. 

Cor. 10, Lines parallel to the same line situated any (477) 
way in space, are parallel to each other. 

For they will be perpendicular to the same plane. 


Fig. 109s. 


PROPOSITION III. 


Tf a plane cut parallel planes, the lines of intersection (478) 
will be parallel. 


For, if the intersections m, n, of the plane, P, 
with the parallel planes, M, N, were not parallel, 
but met on being produced, then would M, N, cut 
each other in the same point, which is contrary to 


the hypothesis, 
Yeas Fig. 110. 


Cor. 1. The segments, as r, s, t, of parallel lines inter- (479) 
cepted by parallel planes, are equal. 


Cor. 2. Conversely, two planes intercepting equal seg- (480) 


248 PARALLEL PLANES, 


ments of three parallel lines not situated in the same plane, are 


- parallel. 


Cor. 3. Parallel planes are everywhere equally distant. (481) 
[Let 7, s, t, be perpendicular to M, N.] 

Cor. 4. Two angles, having their sides parallel and open- (482) 
ing in the same direction, are equal, and their planes are parallel. 

For, let the sides AB, AC, of the angle A, be paral- B 
lel respectively to the sides ad, ac, of the angle a, and 4 
open in the same direction ; draw Bd, Cc, parallel to 
Aa, then will the quadrilaterals, Ab, Ac, Bc, be par- 
allelograms, and the sides of the triangles BAC, bac, 
severally equal,— .. 7 A=a; but Aa =Bb=Ce, 
.. (480) the plane BAC will be parallel to the plane bac. 

Cor. 5. A Dihedral angle, or the angle which one plane (483) 
makes with another, is measured by the inclination of two perpen- 
diculars drawn through the same point in its edge, one in each side, 

For, make A a perpendicular to the planes BAC, bac, then will ’ 
Aa be perpendicular to AB, AC, ab, ac, and the dihedral angle 
BAac will be measured by the plane angle BAC = bac; from which 
it follows that the point A, through which the perpendiculars AB, 
AC, are drawn, may be taken anywhere in the edge, Aa, of the di- 
hedral angle. 

Cor. 6. The segments of lines intercepted by parallel (484) 
planes are proportional. 

For, let ABC, abc, be any lines whatever, pierc- 
ing the parallel planes, P, P, P, in the points A, 
B, C, a, 6, c; and through B draw mBn parallel to 
abc, piercing the planes in m,n. Since A, B, C, : | 
m, n, are in the same plane, and mA parallel to Cn, ie g 

Fig. 1103. 
we have 


AB : BC=mB: Bn=ab: be. 


(4 


e 
Fig. 110e. 


PROPOSITION IV. 


If a line passing through a fixed point revolve in any (485) 
manner so as constantly to intersect two parallel planes, the figures 
thus described will be similar. 


SIMILAR SOLIDS. 249 


For, let VaA, VrX, VyY,..., be positions of 
the revolving line passing through the fixed point, 
V, and piercing the parallel planes in A, a, X, z, 
Y, y, ...; drop the perpendicular VpP, piercing the 
planes in P, p, and join PA, PX, pa, pz. The ra- 
dii vectores PX, pz, have the constant ratio Vp: Vp, 
and Z APX=apz; .°. (426) the plane figures 
AXY, ..., GF, ... 4 are similar, 

Definition. The solid (V, AXY ...) is denominated a Cone, when 
the perimeter AXY ... is wholly curvilinear, and it becomes a Pyr- 
amid when AXY ... is made up of straight lines. The cone is cir- 
cular when its base, AXY ..., is a circle, and right if the perpen- 
dicular fall in the centre of the base. 

Cor. 1. In similar cones [the pyramid is to be included] (486) 
the altitudes, radii vectores, like chords and generating lines for 
corresponding positions, are proportional ; and the bases are as the 
squares of these lines, 

ThowiVPos Vo = PX xcs = chordXyY « chdzy =) VX Vz 
and base(AXY ,..) : bs(azy...) =(PX)? : (pr)? =... 


Cor. 2. If the vertex, V, be carried to an infinite distance, (487) 
the lines Aa, Xz, Yy, ..., will become parallel, and the figure ary 
., will = AXY.,... Under these conditions the solid (AXY ..., 
ary ...) is denominated, a Cylinder when the perimeter AXY ,.. is a 
curve, and a Prism when AXY ... is a polygon; and these magni- 
tudes are said to be right or oblique according as the sides are 
perpendicular to or inclined to the bases. The cylinder and prism 
are also distinguished by their bases. When the base is a parallel- 
ogram it is obvious that all the other faces will be parallelograms 
and those opposite to each other equal, in which case the prism is 
called a Parallelopipedon ; and the Cube is a right parallelopipe- 
don of equal faces. 


x 
Fig. 111. 


SECTION SECOND. 
Surfaces of Solids. 


PROPOSITION I. 


The surface of a Polyhedron, that is, any solid bounded (488) 
by planes, may be found by computing the areas of its several 
faces. 


PROPOSITION II. 


The conver surface of a Right Circular Coneis meas- (489) 
ured by its slant height multiplied into the semicircumference of 
its base. 

Let y be the convex surface included between any 
two positions of its Generatriz, 1, and x the inter- 
cepted portion of the circular base ; then, since y is 
obviously a continuous function of z, giving to y, z, h 
the vanishing increments [k], [h], we have, (311), Fig. 112. 
(148), 


*, (246)* y= fiz; 

Yx— circumference = sl(circumference of base). Q. E. D. 

Cor 1. The circular conical sector is measured by its (490) 
slant height multiplied into half its base. y= diz. 

Cor. 2. The frustrumental surface of the right circular (491) 
cone is measured by its slant height multiplied into the half sum 
of its bases. 

For let VAB, Vad, be conical sectors having the 
same vertical angle, V, and, consequently ABda the 
frustrumental surface in question, we have 


al_\é 


A B 
Fig. 112e. 


* No constant is to be added, since y and x vanish together. 


SURFACES OF REVOLUTION, 251 
ABba = VAB — Vab = VA «- 3AB— Va « 4ab 


= (Va-+aA) - sAB— Va « tab 
=aA +» tAB+ Va « +(AB— abd) 


_ aA(AB + ab) | 
play Spee: 
du ion VA AB d VA—Va_ AB—ab 
Va. “eee al > x eV SA! lui Ghined 
abe aA 
or Va rq: euaey: 


Cor. 3. The convex surface of the Right Circular Cylin- (492) 
der, is equal to its height, multiplied into the circumference 
of its base; for ab becomes = AB. 


Fig. 1123. 
PROPOSITION III. 

If a continuous curve referred to rectangular coérdi- (493) 
nates, revolve about the axis of abscissas, the derivative of the 
surface thus generated, regarded as a function of the correspond- 
ing abscissa, will be equal to the circumference described by the 
ordinate multiplied into the square root of unity increased by the 
square of the derivative of the ordinate also regarded as a func- 
tion of the abscissa. 


For let the surface Z be described by the revolu- M 
tion of any continuous curve, z, around the axis, z, k 
and M, m, k, h, the vanishing increments 
of Z, 2 Y, 23 
then L7= r= [Fz] ’ | 
but (491), M=m[ty+ r(y+4)], 


a 
= (h? + k?)* « (2y+k); 


rs Ea 7 Chea Te Qy= 2ry(1 + Ea i¥ 


as 
or, Z7-r, = 2ry(l+y3_p~)*. Q. E. D. 
It follows that, in order to determine the surface generated by a 
particular curve, we have only to eliminate y and y’ by aid of its 
equation, and to return to the function, 


Q52 SPHERICAL ZONES. 


PROPOSITION IV. 


A Spherical Zone is equal to its altitude multiplied (494) 
into the circumference of the sphere. 

Let Z be a zone generated by the arc, z, of a circle revolving 
about a diameter, which we will assume as the axis of z, the origin 
being at the centre and the radius= 7. We have 

ytot=r, 
CY Ae Ree Lt rit) 
Qyk +k? + 2Qch +h? =0, 


ko ath. 

ho Rytk 

, x Oe a es 
ee Y y= (r2—2?)* a ene y 9 and y a = yp? 

z*\4 seine ae 
2 Zz.n=2y(14+% = Qn(y? + 22)? = Qnr; 
S, ZL = Wrz, 


where there is no constant to be added if we make the surface be- 
gin at the axis of y.. Now let Z,, 2,, Z,, 2, be corresponding values 
of Zand x; we have 
zone(Z, — Z,) = zoneZ, — zoneZ, = (%2—2,) + Te Qr. Q.E. D. 
Cor. 1. A spherical zone is equal to the convex surface (495) 
of the circumscribing cylinder, described by the rev- 
olution of a rectangle with a radius equal to that of ‘ 


the sphere. 


Surface described by S = sur. described by C. 
Fig. 114. 


Cor. 2. The surface of the sphere is equal to the convex (496) 
surface of the circumscribing cylinder, 


' Or, to four Great Circles. aes 
Sur. sphere = 2Z,_, = 4tr’. = . 
Fig. 1149. 


Cor. 3. The surfaces of spheres are to each other as the (497) 
squares of their radii [7], or diameters [2r], or circumferences 
[w+ 2r]. 


RECTANGULAR PARALLELOPIPEDONS, 253 


EXERCISES, 


1°, What will be the expense of gilding a globe 5 ft. in diameter 
at $1‘50 per square foot ? 

2°. What is the surface of the earth and of each of its zones, 
reckoning it as a sphere of 8000 miles in diameter, and the Obliquity 
of the Ecliptic at 23° 28°? 


SECTION THIRD. 
Volumes. 
PROPOSITION I. 


Rectangular Parallelopipedons are to each other as the (498) 
products of their three dimensions, 


First, suppose their corresponding edges OA, A 


OB, OC, oa, 0b, oc, to be commensurable; that 
is, that OA being divided into m equal parts, oa B 
contain an exact number, m’, of the same parts, or 
that : 
g, 


OA =z, 04 = mF, (ir = the com. measure ] 


and OB= ny, 0b = n'y, [y = measure of OB & ob] NS i 
and OC =rz, oc=rz; ia 


then will the partial rectangular parallelopipedons, 4 
formed by passing planes through the points of di- Fig. 115. 
vision parallel to the faces AB, AC, BC, ad, ac, bc, be all equal, 
since any one will be capable of superposition upon any other 
(487). It follows that the solids OABC, oabc, will contain severally — 
mnr, m'n'r’, partial and equal rectangular parallelopipedons, and 


[lt ig 


will consequently be to each other as mur to m'n'r’ ; 


OABC  mnr _ mrenyerz  OA+OB- OC 
cabo Oomnr mae nysrz =§=604+0b+ 0c 


’ 


and the proposition is proved for the case in which the correspond- 
ing edges are commensurable. 

Next, let the dimensions OA, OB, OC, oa, 0b, oc, be any what- 
ever, and put 


254. RIGHT PRISMS. 


OA= A, OB= B, OC=C; oa=a, 0b=b, oc=c. 

Now, if we increase a, b, c, by z, y, Z, so as to become commen- 
surable with A, B, C, and construct the parallelopipedon on 
(a+ 2), (6+), (c+), from what has already been proved, there 
results 


parallelopipedon [(a@ + 2), (6+ y), (¢+2)]_ (a+2)(6+y)(¢+2), 
parallelopipedon [ A, B, C] om ABC 4 
par’dn [a, 8, c] solid [z, y, 2] _ abe ert 
par’dn [A, B,C] ' par’dn [ A, B,C] ~ ABC ABC’ 
.. (63), pard’n [A, B, C] : par’dn [a, b,c] :: ABC : abe. 
Q. E. D. 
Cor. 1. The rectangular parallelopipedon is measured (499) 
by the product of its three dimensions, provided the cube, whose 
edge is the linear unit, be assumed as the unit of solidity. 


For, from par’dn [a, b,.¢)lal) aenb = ¢ = 1, 
we have par’dn [A, B, C]= ABC. 

Cor. 2. The right prism with a right angled triangular (500) 
base, is measured by its base multiplied into its altitude. 

For the diagonal plane divides the rectangular 
parallelopipedon into two rectangular prisms, ca- 
pable of superposition. 


Fig. 1189. 4 
Cor. 3, Any right prism with a triangular base is equal (501) 
to its base multiplied into its altitude. 


For the prism may be split into two, having right 
angled triangles for bases. 


i 

‘ 

i) 

1 

1 
h 
os 
G 


Fig. 1153. 
Cor, 4. Any right prism is measured by the product of (502) 
its base and altitude. 


For the solid may be divided into prisms, having ieee: 
triangular bases. ; . 
Fig. 1154. 
Cor. 5. The right cylinder is measured by the product of (503) 
its base and altitude. 
For (502) is obviously independent of the number and magni- 
tude of the sides, 


PYRAMIDS, 255 


PROPOSITION II. 


The pyramid is equal to one-third of the product of its (504) 
base and altitude, and the cone has a like measure. 


In the first place, suppose the pyramid, y, to have a 
triangular base, z, to which one of the edges, 2, is per- 
pendicular; and, for the purpose of finding the func- 4, 
tion y=fz, give to y, z,z, the corresponding incre- 
ments k, h, 7. Since the prisms constructed with the rh 
altitude A, and upon the bases z, z +7, will be inscrib- YF 
ed in and circumscribe the solid, k, we have Fig. 116. 


kz ae ho 0's sot 9 8s Yap = Bal =z=a7"; (486) 
y=d >. 42° =42 + an* = 472, 
where no constant is to be added, because y,_, =0, and the prop- 
osition is proved for this particular case. 

Next, let a right angled triangle, revolving about its per- 
pendicular, p, and its base, varying in any way whatever, de- 
scribe, the one a cone or pyramid, y, the other its base, x; 
giving to y and xz the vanishing increments, & and A, from 
what has just been proved, we find 


[kK]=4p[h], «. y =4p3 

. Yy = 3pr 3 
and the proposition is demonstrated for all cones and pyramids, in 
which the perpendicular falls within the base. 

Lastly, let the base, u, be any whatever, and the 
perpendicular, p, fall upon its production; then, 
joining the foot of p, with two points of a con- : 
tiguous portion of the perimeter so as to form a 
the base, v, of a second pyramid or cone (p, v), 
we have 


Fig. 1162. 


Fig. 1163. 


cone (p,u-++ v) =4p(u+ rv), 


and cone (p,v) =4pv; 
“ cone (p,u)=4pu. Q. E. D. 


PROPOSITION III 


The Frustrum of a cone or pyramid is measured by (505) 
one-third of the product of its altitude, multiplied into the sum 
of its bases augmented by a mean proportional between them. 


256 VOLUMES. 


Let the base AXY... =A, (fig. 111) 
and 04 <= Bs 
the altitude, VP=2-+4, 
and Vp=2;3 


there results (504), 
solid [(z +a@),A]=3(r+a)A, 
solid [7,B] =4zB ; 
frustrum [A,B] =4(¢+ a)A—47B =t[aA+2(A—B)]; 


A Sai aad 2 ee a PANS 
Ce ge CEE aie 
ee 4 2(A—B) = te 5 (A 4+ BEY(AE — BS); 


“.  frustrum [A,B.a] = 4a[A + CARYL + a (505) 
Cor. The prism or cylinder, whether right or oblique, is (506) 


measured by the product of its base and altitude. 
For B= A, gives ta[A + (AB)! + B] = Aa. 


PROPOSITION IV. 


If a solid, V, be generated by the motion of a plane, U, (507) 
varying according to the law of continuity, and 7 
remaining constantly similar to itself, and per- 
pendicular to the axis of x; then will the deriv- 
ative of V, regarding V as a function of x, be 
equal to the generating plane, U, also regarded 
as a function of x, or 


‘og 4 Ua 7 rT. 


For, from a little reflection, it will be evident that the incremen- 
tal solid [U,U,,h] must be measured as in (505), or that 


Veen = (ee ] = +(U+(UU,)§ + U,] 


=a fet ifr + f(e+h)t?+fle+h)] 
=tfe+fetfrj=fe=U. @Q. E. D. 
Cor, 1. For any solid, generated by the revolution of a (508) 
curve about the axis of z, the ordinate, y, describing the plane, U, 
we have 


ELLIPSOIDs, 257 


Vivre = U= ny? =n( fr)’, 
Cor, 2. For any volume embraced between the surfaces (509) 
described by the revolution of two curves, or the 
two branches of the same curve, U being described 
by the difference of the corresponding ordinates, y, le 
y,, we have 


Vi, = ty? — ry? = 1(y? — yo®) = TY + Y2)(Y — 2). 


Cor. 3. For the ellipsoidal frustrum, estimated from 


Fig. 1172. 


its equator, or from z = 0, we find b 
13 Fig. 1173. 
Vian» — (ate — 40%), (510) 
. 2 4 
since, V{,=17 ST « = (a?—z?), 


Cor. 4, The corresponding frustrum of the circumscribing 
sphere, is 


V, = 1(a®°x — 42°), (511) 
since (510) becomes (511) when 6=a. 
Cor. 5. Prolate Ellipsoid = 2V,_, = 47ab? (512) 
= 3 « 2a » 7h? = #(circumscribed cylinder) PE | 
= 2(inscribed double cone), =. 
Fig. 1174. 
Cor. 6. Sphere, siiue—a= $70? = $ « 2a + Ta? (513) 


= 4(circumscribed cylinder) 
= inscribed double cone) 
=a « 4na? = ta(surface of sphere). 


Scholium, The last relation might have been found by imagin- 
ing the sphere filled with pyramids, having their common vertex 
at its centre, and their bases resting on its surface. 

Cor. 7. The prolate ellipsoid and its circumscribing (514) 
sphere, and their frustrums corresponding to the 
same abscissa, are to each other as the square of 
the minor to the square of the major axis. 


Prolate ellipsoid : sphére, = V, : V,= 6? : a’. 


Fig. 1175. 
Cor. 8. Analogous relations may be found for the Ob- (515) 


17 


258 SIMILAR SOLIDS. 


late Ellipsoid, described by the revolution of the el- 
lipse about its minor axis by changing a into bd, and b 
into a. 
a2 
Thus V,,= 1 « Be 
.. Oblate Ellipsoid = 3 « 2b + ma? = 2(circumse. cyl.) 
= 24+ 2b + ma? = Ainscribed double cone) ; 
Sphere,.ains—5 = 3 ° 2b « 1b? 
.. Oblate Ellipsoid : Inscribed Sphere = V,, : V,,=a? : 5. 


(6° — tzr°), ee =1(0?2 — 42°); 


Fig. 1176. 


Cor. 9. Common Paraboloid V = tpx? = 44a « try? (516) 

= 4(circumscribing cylinder), cree ae 

since Vi= ny? =T « 2pzr; 
Fig. 1177. 


and, for the paraboloid whose equation is 
y?=A,+ Art Agr? +..., 
we have V=T2r(A,+4A ct +4Agr? +...) + Vio. 


213 
PROPOSITION V. 


Similar solids are to each other as the cubes of thetr (517) 
like dimensions. 


We understand by similar solids those in which al] like dimen- 
sions are proportional; and, consequently, the sections through 
~such dimensions similar. 

The proposition becomes evident for the solids already investi- 
gated by making A and B vary as a? in (505), and 6 as a in (512) 
and (515), or by putting A=ca?, B=c,a*, b=c,a, the cs being 
constant, whence 

Similar Cones, Pyramids, and their similar Frustrums, conse- 
quently similar Parallelopipedons, also similar Ellipsoids and 
Spheres 
vary as 4fc+ (ce,)'+c,] « a3, 403 « a, 
or as a°, or as the cubes of any like dimensions. 


Next let any two similar perimeters, whether rectilin- (fig. 99.) 
ear or curvilinear, be similarly placed and revolve about any line 
OaA of corresponding radii vectores as an axis of x, the origin be- 
ing at O; the ordinates y, y,, of the extremities P, p, of any other 
corresponding radii vectores, will describe planes U, U,, terminating 


EXERCISES, 259 


like frustrums of the two solids estimated from O where z = 0, and 
this whether the radii y, y,, remain constant for a given value of 
x or vary in any way whatever, that is, whether the perimeters 
ABC ..., abc ..., continue of the same magnitude or be variable, 
only that they preserve their similarity. Therefore we have 

te Voreth, 6 eee nti ee as 

“. (507) eV 2 Veeifrustrum[ 7) «ff 02)] = 42° :,403-O0P? : On’, 
and the proposition is proved for this more general case. 

Lastly, let the similar solids be any whatever, and assume any 
two like diameters for the axes of x, 2, the origins dividing them 
proportionally. It follows from the definition of similar solids 
given above, that the generating planes U, U,, perpendicular to z, 
Zz, Will have corresponding positions in their respective solids V, 
V., when any diameter in U is to a like diameter in U, as the ab- 
scissa 2 terminating in U is to the abscissa x, terminating in U,, and 
that U and U, will be similar figures ; 


s Oe eh s see) es 
But from (300) it is manifest that (507) is applicable in this case 
also ; 
VU 28a Uy 0, "> 2. 
od) Vs V,2if 0) Uy SH 40% beg 2? : 2p. Q, ED. 


EXERCIS&S. 


1°. A cylindrical cistern, capped with a hemispherical dome, is 
15 feet deep and 10 in diameter, Required its capacity. 

2°, A hollow cylinder, the side of which is one inch thick, is set 
into a cubical box, touching its sides and equalling it in height, 
and within the cylinder is placed a hollow sphere also an inch in 
thickness and tangent to the cylinder, What is the capacity of the 
sphere, it requiring just 10 gallons of water to fill the space between 
the box and the cylinder ? 

3°, What is the capacity of a cask, regarded as a frustrum of an 
ellipsoid, the bung diameter being 30 inches, the head diameters 
25 each, and the length 40 inches; and how much will it differ from 
its inscribed double conical frustrum ? 

4°, What is the capacity of a paraboloidal cistern, having the 
dimensions in 1°? 


260 EXERCISES. 


5°, Wishing to ascertain the weight of a marble column 30 feet 
high, I take the semidiameters at the elevations 0, 10, 20, 30 feet, 
and find them to be 3, 4, 3, 2, feet; the specific gravity of marble 
is 2°7, water being 1, and a cubic foot of water weighs 1000 oz. 
avoirdupois. What is the weight? 

6°, The earth may be regarded as an oblate spheroid, generated 
by the revolution of an ellipse about its shorter diameter of 
7899‘170 miles, while the equatorial diameter is 7925‘648, accord- 
ing to Sir J. F. W. Herschel. Required the excess of volume over 
the inscribed sphere, and the quantity of water on the surface; al- 
lowing the sea to be to the land as 4 to 3, and its mean depth to 
be 22 miles. 

7°, What must be the dimensions of a tub to hold 10 cubic feet, 
the depth and two diameters being as the numbers 4, 5, 62 

8°. What must be the dimensions of a paraboloidal kettle to 
contain 12 gallons, the diameter across the top being to the depth 
as 7 to 8? 


BOOK SECOND. 


SPHERICAL GEOMETRY. 


SECTION FIRST. 
Spherical Trigonometry. 
PROPOSITION I. 


A Spuere, being a solid bounded by a curve surface (518) 
everywhere equally distant from a point within, called the cen- 
tre, may be generated by the revolution of a semicircle about tts 
diameter. 

For, let the semicircumference, PeEP’, 
revolve about its diameter, PoOP’; it 
follows that every point, e, in PeEP’ will, 
while describing the circumference, ee’e”’ 
.. » Maintain a constant distance, 

Oe=O0el=VE =... =OP'= OP 
= OR = OF = ..., 


a2 


from the centre, O; whence the surface 1S 
described will be that of a sphere. T 
Fig. 118. 


Cor. 1. The radius, OF, perpendicular to the axis, (519) 
POP’, will describe a Great Circle, the circumference of which, 
EEE’ ... , denominated the Equator, will be everywhere 90° dis- 
tant from its poles, P, P’. 

Cor. 2. Any other perpendicular, oe, will describe a (520) 
Small Circle, also perpendicular to the same axis, and having the 
same poles, P, P’. 

Cor. 3, Every section of a sphere by a plane is acircle. (521) 

Cor, 4, The perpendicular, P’T, through the extremity (522) 


262 LUNES. 


of the diameter, POP’, will generate a plane (P’,TT'T” ... ) tan- 
gent to the sphere. Hence, a plane passing through the extremity 
of any radius, and perpendicular to it, is tangent to the sphere ; 
and, conversely, a tangent plane is perpendicular to the radius, 
drawn to the point of contact. 

Cor. 5. All great circles mutually bisect each other; as (523) 
the Meridians, PEP’, PE'P’, since they have a common diameter, 
REY, 

Cor. 6. Every great circle bisects the sphere. (524) 

Cor. 7. A small circle divides the sphere into unequal (525) 
parts, and is less the more distant it is from the centre. 

Cor. 8. A Lune, or the spherical surface embraced by (526) 
two meridians, is to the whole surface of the sphere as its equato- 
rial arc to the total equator; thus, 

Lune PEP’EP : Sph. Surface = arcEE’ : 360°, 
or Lune, >. 40r? 2° eB Onesie 
e° é = eq.are. 
Lune. = 695 * al E = rad.sph. 

Cor. 9. The Spherical Wedge or Ungula, PEP'E‘O, is (527) 

to the whole sphere as its equatorial arc is to the total equator ; 
Ungula,, 2.4 o7r* = e°_: 860°, . 

Cor. 10. Every meridian is perpendicular to its equator; (528) 
as PE'P’ to EEE’ .... 

Cor. 11. A Spherical Angle is identical with that em- (529) 
braced by the tangents to its sides, and is measured by the arc of 
its equator intercepted by these sides; as the angle EP’E’ = TP’T’ 
measured by EE’, 

Definition. A Spherical Triangle is the surface em- (530) 
braced by three arcs of great circles, 


(526) 


PROPOSITION II. 


A spherical triangle ts equal to the sum of its three (531) 
angles diminished by a semicircumference, multiplied into the 
square of the radius of the sphere. 

Produce one side, AB, of the spherical triangle, 
T, so as to form the circumference, ABPQ, inter- 
sected in P and Q by the productions of the other 
sides, AC, BC; then, by the addition of the trian- 
gles, X, Y, Z, to T, there will be formed the lunes Fig. 119 


TRIANGLE, 263 


T + X, T+, having the angles A, B, and to the surface T+ Z, 
which may be readily shown by (523) to be equal to a lune with 
the angle, C ;* and, from (526), there results, 


A? , ne Ste 
P+ X= oe 3 tr, T+ Y= 


» SPO mrt PLEX VER) a eT po 
A° + B° + C° — 180° 
180° 

Cor. 1. The sum of the three angles of a spherical tri- (532) 

angle is always greater than two right angles and less than six ; since 
for the existence of T’ we must obviously have 

Pe: 

At BHO 7 pet 

Cor. 2. Similar spherical triangles are to each other as (533) 

the squares of the radii of their respective spheres, or as the squares 

of their homologous sides. 


fo) 
Tr, Py tyke Tr? ; 


2.12 s otr?=(A+B+C—n)r*, (531) 


Cor. 3. A spherical polygon is measured by the sum of (534) 
its angles diminished by as many semicircumferences as it has sides 
save two, multiplied into the square of the radius of the sphere. 

For the number of sides being n, the polygon may be divided 
into (n— 2) triangles, T, T., T;, ..., whose angles A, B,C; Ay 
B,, C,; As, B;, Cy;..., make up the angles of the polygon, and 
there results 


(T+ T,+ Ty; +... )ne = (A+B+ C—t)+(A-+ B+ C,—t)... Jr’, 
or P,=[S —(n—2)r] + r*?, S=sum of angles. 


* For, producing the arcs CP, CQ, till they meet in R, 
and drawing the diameters (?) AOP, BOQ, COR, we 
have only to show that the triangle PRQ, which com- 
pletes the lune CPRQ, is equalto 4 ACB. In order to 
this, take the point S, equally distant from A, B, C, or 
the pole of the small circle passing through these points, 
and draw the diameter SOT; then the arcs SA, SB, SC, 
TP, TQ, TR, will be all equal; also(?) Z ASB=PTQ, 
BSC=QTR, ASC=PTR, and it may therefore be : 
shown by superposition that Fig. 1192. 


A QTP = ASB, RTQ =BSC, RTP = ASC; 
APQR=QTP-+RTQ—RTP 
=ASB+ BSC —ASC=ABC. Q. E. D. 


264 FUNDAMENTAL THEOREM 


PROPOSITION III. 


The cosine of any side of a spherical triangle is equal (535) 
to the product of the cosines of the other two sides increased by 
that of their sines multiplied into the cosine of the angle opposite 
the first-mentioned side. 


For, denoting the angles of 
the triangle by A, B, C, and the 
sides respectively opposite by 
a, b, c, let the radius of the 
sphere OA = OB = OC, be ta- 
ken for unity, and draw the 
tangents, tand, tanc, terminat- - 
ing in the productions of OC, 
OB, at Q, P, and join PQ=m; then will OQ =secd, OP =secc, 
and the spherical angle A will = Z PAQ, also Z BOC will be meas- 
ured by a, (?), and from (388) there results, 


Fig. 120. 


sec*b ++ sec*c = m? + 2secb secc cosa, 
and tan?b + tan*c = m?-+- 2tand tanc cosA; 


1+ 1 = 2secb secc cosa — 2tanb tanc cosA, 


1 1 sind sinc 
or = —— e —— e cosa — ——  - —— « cosA; 
cosh cosc cosh cose 
whence cosa = cosb cosc + sind sinc cosA ; 
SO cosb = cosa cosc + sina sinc cosB, (535) 
and cosc = cosa cosb + sina sinb cosC. 


Cor. 1. If two spherical triangles have two sides of the (536) 
one equal to two sides of the other each to each but the included 
angles unequal, the remaining sides will be unequal and the greater 
side will be opposite the greater angle; for, as A increases, cosA 
decreases and, consequently, cosa decreases or a increases, there- 
fore A and a increase or decrease together. 


Cor. 2. Two spherical triangles are equal: (537) 
1°, When two sides and the included angle of the one are re- 
spectively equal to the two sides and the included angle of the oth- — 
er; and therefore, 


FOR A SIDE, 265 


2°, When the sides of the one are severally § 4 
equal to the sides of the other. 


Far, 1°, if b= 65.66 yA til 


~~ 


then a=a;\. B=B,C=C'’ B B! 
. tr(ABC) =(A' BC) ; 
‘ ley i sig mer ys Cc’ 
2°, if gas 2 > Roe bon Fig. 1209. 
then A= A’; .. tr(ABC = (A'BC)). 


Cor. 3. The arc joining the vertex and the middle point (538) 
of the base of an isosceles spherical triangle, is perpendicular to the 
base, and bisects the vertical angle. 

Cor. 4. We may adapt (535), which is the fundamental theorem 
of spherical trigonometry, to logarithmic computation, when a side 
is required, by putting 


sinb cos A ge eagee cosb ; 
cos 
for then (535) becomes 


; sinu 
cosa = cosb cosc + cosb sinc « 
cos 


Cc Ss e e 
= —— (cose cosu + sinc sinw) ; 
cosu 


and there results 
cosb cos(c—u) 


tanw = tand cos.A4, cose = ——__—____—_; 
Costu 
cosa cos(c—v 
sO tanv = tana cosB, cosb = Ae Got (539) 
cosv 
cosa cos(b — w) 
and  tanw = tana cosC, cosc = ——_____——- 


CcOsw 


Cor. 5. When an angle is required, the transformations of 
Plane Trigonometry may be imitated, and we have 
2sin*4+A = 1—cosA 
my, cosa — cosb cosc 
a } - sind sinc 
cosh cosc + sind sinc — cosa 
ae sind sinc 
cos(b — c) — cosa 
mt sinb sinc 
Qsint(a +b —c) « sint(a—b+c) 
i sind sinc 


266 FOR AN ANGLE. 


dia = Ee sint(a+b—c) « ae 
sind sinc 
sin(h sin(h —c)]4. 
= [Ree = sinc sae |? ; 
nt B= ps (b+ a—c) + sint(b+c— ai? 
sina sinc (540) 
sin(d — a) sin(h— c) 
ie oy Spore pee sind sinc th 
and — sink C= hese (c-+a— 6) « sing(e-+b — a+ 
sind sinb 
sf eget — a) sin(h— b)]4 
i sina sind | : 
And by a like process we shall find 
eas Tae sin(h — al + 
ah sind sinc 
sink sin(h—b)]4 
sing sinc 


sink sin(h — c)]+ 
sing sinb 


(541) 


cost B = [ 


cost C= [ 


sin(h — b)sin(h — c)q2 7 

sink sin(h—a) J , 
sin(h — a)sin(h — c)74 

sinh sin(h —d) ’ (542) 
sin(h — a)sin(h — he 


sink sin(h — c) 


oe tant A= = 


tan4_B = [ 


tant C = [= 


Cor. 6. The sum of any two sides of a spherical triangle (548) 
is greater than the third side. 

Cor.7. If from any point within a spherical triangle, arcs (544) 
of great circles be drawn to the extremities of either side, the sum 
of the including sides will be. greater than the sum of the included 
arcs. [See Plane Geometry. ] 

Cor. 8. A chain of spherical arcs is less the nearer it lies (545) 
to the arc joining its extremities. 

Cor. 9. The arc of a great circle is the shortest distance (546) 
from point to point on a spherical surface. 

Cor. 10. The sum of any two of the plane angles that (547) 
make up a solid angle, is greater than the third angle. 

For, if about the solid angle, O, a sphere be described, 


REGULAR POLYHEDRONS, 267 


the plane angles AOB, BOC, AOC, 
will be measured by AB, BC, AC, 
the arcs of the spherical triangle ABC, 


Cor. 11. The sum of all the plane angles that make up (548) 
a solid angle, is less than four right angles. 

For, let a, b, c, ... , 7, k, 1, be the arcs of the spherical polygon 
by which the plane angles are measured, and produce aq, ¢, till 
they meet, forming the arcs,a+m, c+n; the new polygon 
whose sides are a+ m,c-+n, d,..., 1, will have a greater peri- 
meter, consisting, however, of a number of sides less by one, than 
the original polygon, And this reduction may be continued till a 
triangle is obtained whose perimeter, less than a circumference, 
will be greater than that of the polygon. 


Cor. 12. There can exist only (549) 


Five Regular Polyhedrons. 


Three—Tetrahedron, Octahedron, Icosahedron, whose faces are 
equilateral triangles, [360° : 60° =6]. 

One—Hexahedron, squares, [360° : 90° = 4]. 

One—Dodecahedron, regular pentagons, 


These figures may be formed of | | Os 


pasteboard, the lines of folding being 
cut half through 


: Fig. 1203. 
Cor. 13. If the three plane angles that constitute two (550) 
solid angles be respectively equal, the homologous planes will be 
equally inclined to each other. 


PROPOSITION IV. 


In any spherical triangle the sines of the sides are to (551) 
each other as the sines of the angles respectively opposite. 

Taking the double products of (540) and (541), observing that 
Qsint A costA = sinA, we have 


268 SIDES AND ANGLES. 


sinA = : : 3 
sind sinc 

cigs ain inh — a) sini oa) oal oma (552) 
sing sinc « 

4 eg A VP kde oe 5s | 

Tat aa ee eS a a ee 

sina sinb 
es YL ee Me: Le ee 


‘* sinB sind’ sinC sinc’ sinC sinc’ 

Cor. 1. In a spherical triangle the greater side is opposite (553) 
the greater angle, and v. v. 

Cor. 2. The perpendicular arc is the shortest distance (554) 
from any given point to an are also given in position. 

Cor. 3. The angles opposite the equal sides of an isos- (555) 
celes spherical triangle, are equal, and vice versd. 

Cor, 4, An equilateral spherical triangle is equiangular, (556) 
and v. v. 


PROPOSITION V. 


To eliminate from equations (535) the cosine of each side in 
succession. 


Substituting the value of cosc from the third in the second, we 
find 
cosb = cosa(cosa cosb-+ sina sinb cosC) + sina sinc cosB 
= cos*a cosb-+ sina cosa sinb cosC' + sina sinc cosB; 
“. (1 —cos*a)cosb = sina cosa sinb cosC’+ sina sine cosB 
or sin*a cosb=sina cosa sinb cosC + sina sinc cosB, 
and sina cosb = cosa sind cosC’+ sinc cosB, 7 
so sina cosc = cosa sinc cosbh + sinb cosC, 
sinb cosc = cosd sinc cosA-+ sina cosC ; 
sinb cosa = cosb sina cosC +- sinc cosA, (557) 
sinc cosa = cosc sina cosB-+ sind cosA, 
sinc cosb = cosc sinb cosA + sina cosB. | 


Vote. Instead of going through independent operations, all the 
above forms may be obtained from the first by a simple change of 
letters ; thus, changing 0b, B, into c, C, and vice versd, the first 
becomes the second, which in turn gives the third by commuting 
a, A, and b, B, and so on. 


FORMS, 269 


PROPOSITION VI. 


From (557) to eliminate a sine, in order that no more than two 
sides may be embraced in each equation. 


sinc * sinC sinbsinC 

From (551) we have —- = ——., or sinc = —- which sub- 

Scoae, sind sinB’ snB ” ee 

stituted in the first of (657), gives 
. ‘ sinb sinC 
sina cosb = cosa sinb cos C + ———.— « cosB, 
sinB 

: cosh cosB 

> sing e ——=cosa cosC : e sinC; 
* sind a sinB : 

ae sina cotb = cosa cosC' + cotB sinC, 


sind cotc = cosa cosB+ cotC sinB, 
sinb cotc = cosb cosA + cotC sinA ; 
sind cota = cosdb cosC' + cotA sinC, 
sinc cota = cosc cosB + cotA sin B, | 
sinc cotb = cosc cosA -++ cotB sinA. 


PROPOSITION VII. 


From (558) to eliminate a side. 
From the fourth and first by aid of (551) we have 


cosh : 
cota= = al" cosC + cotA sinC « 
sind 


fi 


sind 
sinA 


= coth co cotA sinC « ———_ 
Y sor sina sin B’ 


‘ 1 
and cotb = cota cosC + cotB sinC + ——_; 
sina 


; 1 
cota = cota cos*C + cotB sinC cosC « 
sina 


‘ sinA 
cotA sin C @ a 
= sin A sinB’ 
; ; cotBsinCcosC | cotAsinA sinC 
and cota(1 — cos*C) = ———_— SACRE tog ae 
sind sing sin.b 
cosa, cosB sinC cosC . cosA_ sinAsin’ 
or - e sin? C = —_ « ——_ + —— a ——— 3 
sind sin PB sina sinA sina sinB 


.. sinB sinC cosa =cosB cosC + cos; 


270 POLAR TRIANGLES. 


.. cos(180°—A) = cosB cosC' +sin B sinC cos(180°—a), 
cos(180°—B) = cosA cosC +sinA sinC’ cos(180°—S), 0 
cos(180° —C) = cosA cosB +sinA sinB cos(180°—c). 


Cor. 1. If in these equations we substitute 


A = 180° —a, a = 180° — A, 
B= 180° —b, b = 180° — B, 
C= 180° — ¢; c = 180° — 0; 


there results, 
cosa = cosb cose + sinb sine cosA, 
cos) = cosa cos¢ + sina sing cosB, 
cos¢ = cosa cosh -+ sina sinb cosO ; 
which are of the same form with (535), and therefore belong to a 
triangle whose sides are a, b, ¢, and angles A, B, 0; the relation of 
these two triangles, ABC, ABC, is 
A+a=180°=A-<a, 
B+b=180°=8-+4, {00 
C+e=180°=C+c; 
that is, the sides of the one are supplementary to the angles of the 
other; hence they are denominated Supplemental Triangles, 
more commonly known as Polar, from their geometrical construc- 
tion, which is as follows. 

About the angles A, B, C, as poles, with a 
radius of 90°, describe the arcs BC =a, AC=D, 
AB =¢, forming the spherical triangle ABC, polar 
to ABC; U, being the intersection of BC, AC, equa- 
tors to A, B, is 90° distant from A, B, and there- 
fore the pole of AB, and in the same way A, B, 
are shown to be the poles of BC, AC; .°. if two 
sides, as AB, AC, be produced to intersect BC in m, n, we find 

Bm + mn = 90°, mn+ nl =90°; 
mn + (Bm + mn-+ nf) = 180°, 


or A+a= 180°; [mn = measure of A] 
so B+hb=180°, 
and C+¢6= 180°; 
again A+a=pq+ BC=pC-+ Bg = 180°, 
B+6= 180°, 
(+c= 180°. 


Cor. 2. Instead of transforming (559) we may apply (560) to the 
polar triangle of (540), and 


FORMS. 271 


anak ze + ao | 4 
sin) sin¢ 
becomes sin4(180° — a) = 
ee 2 ee ee ees + 
—..... -in(180° — B) sin(180°—C) i 
cos}(A + B— C) cost(A+ C— = 4 
ae Hen dm Ceres IT 
or, putting H=3(A+B-+ C), 


cosja = [= B) cos(H — eats , 


or costa = [ 


inB sinC 
$e oe eae 561 
abe [ sinA. sinC J (561) 
_ [-cos( H — A) cos( H — B)q4 
conte =| aR | 
tid Oot bie cos Hf cee z 
bis a sinB sin? . 

‘ —cosH cos(H— B)]4 
si gine oe cae | (562) 

ett) Ee cosH cos(H — a t. 
i te sinA sinB ; 


—cosH cos(H — A) i}? 
= ere 


— cosH cos(H — B) |? 
tantd = care ae ool (563) 


— cosH cos(H — C) a 
Bi 4e=| - os(H — A) saree B) 


Cor. 3. Applying to (559) a process like that for (539), we find 
tan U=tan B cos(180°—a), cou(te0ee=Aj eee cos( C—U) 


cos U ; 
A CV 
tan V=tanA cos(180°—S), cos(180°—B)=— i ) (564) 
cos Acos( B— W 
tan W=tan A cos(180°—c), cos(180°—C)= Samed 


Cor. 4. Forms (383), (384), may be frequently applied to (565) 
(535), (557), (558), and (559), with advantage. 


272 NAPIER’S ANALOGIES, 


PROPOSITION VIII. 
Napier’s Analogies. 
cost(a-+ 5) : cost(a —b) = cottC : tant(A-+ B), (566) 
sint(a +6) : sint(a—b) =cotsC : tant(A — B) ; 
cos}(A+ B) : cost(A — B) = tante : tand(a+d), (567) 
sint(A-+ B) : sint( 4 — B) =tante : tand(a — dD). 
Adding the first and fourth of (557), we get 
sin(a-+ 6) =sin(a+ 0) cosC + (cosA + cosB) sine, 
or (cosA + cosB) sinc = sin(a + b) (1 —cosC); 
but (551), sinA sinc = sina sinC, 
sinB sinc = sind sinC ; 
(sinA + sinB) sinc = (sina+ sind) sinC, 
(sinA — sinB) sinc = (sina — sind) sinC ; 
sinA+sinB _ sina+ sind sinC 
cosA-+cosB sin(a+b) \1—cosC’ 
sinA —sinB _ sina — sind sinC 
cosA+cosB sin(a+6) 1—cosC’ 
and by (335), (329), (321), (330), (321), (336),_ ,, the last two forms 


become 


cost(a — b) 


1 anid ee nel: 
tant, A +B) — Se Th oes ee 
any( = Page cree 8 « € 2 9 
* _ cosd(A — B) gota 
.*. (560) tant(a+b)= cosita TOBY tante, 
Ah och, Sin A (567) 
anz(a — hen AaB) e tangc. 


The above proportions (566), (567), should be expressed in 
words, and committed to memory; since they make known by 
easy logarithmic operations, 1°, when two sides and. the included 
angle are given, the remaining angles; and, 2°, when two angles 
and the included side are the data, the remaining sides, 


NAPIER’S RULES. 273 


PROPOSITION IX. 
Napier’s Rules. 
If, in a right angled spherical triangle, we denominate 


CIRCULAR PARTS, 


The side opposite the right angle with the including angles, 
and the complements of the other two sides, then 
Cos. Mid. Part = Product of Sines of Opp. Parts (568) 
= Product of Cotangents of Adj. Parts, 
observing that each of the five parts will be adjacent to two and 
opposite to two, no account being made of the right angle. 

In order to demonstrate this proposition, which is a most remark- 
able example of artificial memory, it is only necessary to make 
A = 90° and seek among the preceding forms, values for 

6B C9009. b=! by 90? —~ 0.2 'c, : 
we find 
cosa = cosb cosc, or cos.b cosC =sinB sinC cosa ; 
cosB = sinC cosd, or sinc cosa = cose sina cosB ; 
cos C'=sinB cose, or sind cosa = cosb sina cosC ; 
sind = sina sinB,or sinb cotc = cotC ; 
sinc = sina sinC, or sinc cotb =cotB ; 
and these reduce to 
cosa = sind, sinc, = cotB cotC, 
cosB = sinb, sin C = cota cotc,, 


cosC = sinc, sinB = cota cotb,, (568). 
cosh, = sina sinB = cotc, cotC, 
cosc, = sina sinC’ = cotb, cotB. | 


Cor. 1. Napier’s Rules may be employed in solving (569) 
Quadrantal Triangles, by observing that the circular parts will 
be the supplement of the angle opposite the quadrant, with the 
sides adjacent, and the complements of the other two angles. 

For, let @ = quadrant = 90°, 
then (560), A = 180° —a=90°, 
and the triangle ABC, polar to ABC, being right angled at A, 
gives (568) 

cosa = sinh, sine, = cotB cot, 
.*, (5660) cos( 180° — A) = sin[ 90° — (180° — B) ]sin( C — 90) =..., 
or, putting 180° — A= A,, &c., and reducing, 
* May be read the complement of b, or 6 complement. 


18 


274 BOWDITCH’S RULES. 


cosA, = sinB, sin C= cotB cotC ; 
SO cosb = sinc sinB,= cotA, cotC,, 
cosc = sind sinC’, = cotA, cotB,, > (569) 
cosB, = sinA, sind = cotC, cotc, 
cos C’.= sin A, sinc = cotB, cotd. 

It will be observed that the quadrant, a, is regarded as not sepa- 
rating B, C. 

For illustration, let it be required to find B when a and c are 
given, and b= 90°. 

We have cos(180° — B) = cosB,= cota cote. 

Cor, 2. Napier’s Rules may be employed in the solution (570) 
of Isosceles Spherical Triangles, by observing that the circular 
parts will be the equal side, equal angle, half unequal angle, and 
complement of half unequal side. 

For, letc =a; then joining B and the middle point of b, we 
shall have formed two right angled triangles, in each of which the 
parts will be (538), (568), 

a, A,+B, (45). ; 


cosa = cotA cotsB, 7 
cosA = cota cot(4d),, 
cos(4b), = sina sintB, + (570) 
cos$B=sinA sin(4d),. | 


Cor. 3. Napier’s Rules may be employed in the solution (571) 
of Oblique Triangles, by dropping a perpendicular so that two 
of the known parts shall fall in one of the right angled triangles 
thus formed; there will result also, 


Bowditch’s Rules. 


1°, The cosines of the parts opposite the perpendicular, are pro- 
portional to the sines of those adjacent. 

2°, The cosines of the parts adjacent to the perpendicular, are 
proportional to the cotangents of the parts opposite. 

For, let the angle, C, be separated into the parts, 
M, N, by the perpendicular, p, dividing the oppo- 
site side, c, into the corresponding parts, m, n, 
subjacent to a, b, and we find, 


Fig. 1212. 


cosa = sinp, sinm,, cosb = sinp, sinn, ; (1°) 
cosB = sinp, sinM, cosa =sinp, sinN ; 
cosm, = cotp, cotB, cosn, = cotp, cotA ; (2°) 
cosM = cotp, cota, cosN = cotp, cotd. 


CASES IN SPHERICAL TRIGONOMETRY. 275 


Cases in Spherical Trigonometry. 


Given. Forms for Solution. 


Pi 


I, |'Two sides and included angle. | (535) ; (539) ; (566) ; (571). 


Il. |Twoanglesand included side. | (559) ; (564) ; (567) ; (571). 
535) and (384) ; (551), [?]. 
= |T'wo angles and opposite side, |(559) and (384) ; (551). 


( 
IL. |'Two angles and included side. |( 
ILI.|T'wo sides and opposite angle. | ( 
( 
(8 


V. |Three sides. 35); (540); (541); (542); (552). 
VI/Three angles. (559) ; (561); (562); (563). 
| Right Triangles. (668). 
| Quadrantal Triangles. (569). 

[Isosceles Triangles. (570). 


Scholium J. There will bea choice in forms, not only on 
account of logarithmic operations, but also for the purpose of 
avoiding the cosines of very small ares, and the sines of those dif- 
fering little from a quadrant. 

Scholium II. Problems pertaining to spherical trigonometry 
will generally find their easiest solution by constructing the trian- 
gle so that one of its angles shall be at the pole of the sphere. 


Latitudes and Longitudes. 


Places. Latitude. Longitude. 


Boston (State House), Mass., 4Q°Q1'2257'N,| 71° 4° OW. 
Chicago, II1., 42 0 O N.| 87 35°) OW. 
Canton, China, 23 8 9 N./113 1654 E. 
Cape Good Hope (Obs.), Africa, 33 56 3 S.) 18 2845 E. 
Cape Horn, S. America, 55 58 41 =S.| 67 1053 W. 
Cincinnati (Fort Washington), Ohio,)39 5 54 N.| 84 27 OW. 
Greenwich (Obs.), Eng., 51 28 39 N.| O 0 0 

New York (City Hall), N. Y., 40 42 40 N. 74 1 SW. 
Paris, (Obs.), France, 48 50 13 N.| 2 2024 E, 
Philadelphia (Ind’ce Hall), Pa., 39 56 59 N.| 75 954 W. 
Rome (Roman Col.), Italy, 41 53 52 N,| 12 2840 E. 
Washington (Capitol), D. C., 38 53 23 N.| 77 1 24W. 


276 EXERCISES, 


1°, Required the distance and direction from New York to 
Greenwich. [69*2 miles to 1°.] 
Polar distance of New York = 49° 17’ 20", 
Polar distance of Greenwich = 38° 31 21"; 


Difference of longitude dt e aahiae Saas = die 
the problem, therefore, belongs to Cases I., III. 

(539), tanu = tan49° 17’ 20” 0:065262 

<x cos74° 1’ 8" 1439838 

u=17° 44 34” 1505100 

cosa = cos49° 17’ 20" 1814411 

< cos20° 46' 47” 1970789 

: cosl7° 44’ 34” 1978835 

a = 3473 miles, 1'806365 

(551), sind50° 11’ 17” 1‘885446 

: sin74° 1’ 8” 1*982883 

= sin38° 31’ 21” 1°794363 

: sind1°12'43" 1891800 


Solve the same by (566), also by (571). 

2°. Required the distances between New York, Cincinnati and 
Washington, the angles of the spherical triangle thus constructed, 
and its area. 

3°. Required the distance and bearing of Chicago from Boston, 
reckoning the latitudes of both places at 42°. 

4°, Required the direction and distance from Greenwich to 
Quito, the latter place being under the equator, and having 79° W. 
lon., nearly. 

5°. Required the breadth of South America between Cape 
Blanco and Cape St. Roque, reckoning both to have 4° 30’ S. lat., 
and a difference of longitude = 46°, 

6°. To reduce an angle to the horizon, (572) 

Around the point of observation, O, imagine a 
sphere to be described with radius = 1; then will 
one of its arcs, a, be the measure of the inclined 
angle, SOS’, and the other two, 4, c, of the angles, 
S'OP, SOP, made with the vertical, OP ; whence 
the spherical angle, A, opposite a, will be equal to 
the horizontal angle required. 


EXERCISES, 277 


For example, let a = 367°, b= 110°, c= '75'25°; then will A 
be found = 12° 0' 20”. 

7°. Through any two given points and a third upon the surface of 
a sphere, which do not lie in the circumference of the same great ) 
circle, there may be made to pass two equal and parallel small cir- 
cles; that is, one of them through the first two given points, and 
the other through the third given point; and every spherical are 
which is terminated by these circles shall be bisected by the cir- 
cumference of the great circle, to which they are parallel.* 

8°. If there be two equal and parallel small circles, and if a 
great circle meets one of them in any point, it will meet the other 
in the opposite extremity of the diameter which passes through 
that point. 

9°. If a great circle cuts one of two equal and parallel small 
circles, it will cut the other likewise; also, if it touches one of 
them, it will touch the other likewise. 

10°. Lunular portions of surface, which are contained by equal 
spherical arcs with the arcs of equal small circles of the same 
sphere, are equal to one another; so also are the pyramidal 
solids, which have these portions for their bases, and their com- 
mon vertex in the centre of the sphere. 

11°. Spherical triangles, which stand upon the same base and 
between the same equal and parallel small circles, are equal to one 
another. 

12°. If equal triangles, ABC, EBC, stand upon the same base, 
BC, and the same side of it, the points, A, E, and B, C, lie in the 
circumference of two equal and parallel small circles, 

13°. Of equal spherical triangles upon the same base, the isos- 
celes has the least perimeter. 

14°, Of all triangles which are upon the same base, and have 
equal perimeters, the isosceles has the greatest area. 

15°. If two spherical triangles have two sides of the one equal 
to the two sides of the other, each to each, and the angle which is 
contained by the two sides of the first equal to the sum of the other 
two angles of that triangle, but the angle which is contained by the 
two sides of the other not so; the first triangle shall be greater 
than the other. 

16°. Two given finite spherical arcs, together with a third inde- 


* This and the following exercises, drawn from the Library of Useful Knowledge, 
may be performed at the student’s leisure. 


278 EXERCISES. 


finite, inclose the greatest surface possible, when placed so that the 
included angle may be equal to the sum of the other two angles of 
the triangle, 

17°. In a triangle, ABC, which has one of its angles, ABC, 
equal to the sum of the other two, the containing sides, AB, BC, 
are together less than a semicircumference. 

18°, Of all spherical polygons, contained by the same given 
sides, that one contains the greatest portion of the spherical surface 
which has all its angles in the circumference of a circle. 

19°. A circle includes a greater portion of the spherical surface 
than any spherical polygon of the same perimeter. 

20°. The lunular surface, which is included by a spherical arc, 
and a small are, is greater than any other surface which is included 
by the same perimeter, of which the same spherical arc is a part. 

21°, Of all spherical polygons having the same number of sides 
and the same perimeter, the greatest is that which has all its sides 
equal and all its angles equal. 

22°. Spherical pyramids, which stand upon equal bases, are 
equal to one another; so, likewise, are their solid angles. 

23°. Any two spherical pyramids are to one another as their 
bases, and the solid angles of the pyramids are to one another in 
the same ratio. 

24°. Every spherical pyramid is equal to the third part of the 
product of its base and the radius of the sphere. 

25°, Every solid angle is measured by the spherical surface 
which is described with a given radius about the angular point, and 
intercepted between its planes. 

26°. To find the diameter of a given sphere.* 

27°. To find the quadrant of a great circle. 

28°, Any point being given upon the surface of a sphere, to find 
the opposite extremity of the diameter which passes through that 
point. 

29°, ‘To join two given points upon the surface of a sphere. 

30°, A spherical are being given, to complete the great circle of 
which it is a part. 


* In executing this and the following problems, it is not permitted to employ any 
thing like a flexible ruler or straightedge, but the student is supposed to be furnished 
simply with a pair of compasses of such construction as to be capable of embracing 
the extremities of any arc not greater than a quadrant. A Spherical Blackboard 
would be found useful. 


EXERCISES, 279 


31°. To bisect a given spherical arc, 

32°. To draw a spherical arc which shall bisect a given spherical 
are at right angles. 

33°. ‘lo draw an are which shall be perpendicular to a given 
spherical arc, from a given point in the same. 

34°, To draw an are which shall be perpendicular to a given 
spherical arc, from a given point without it. 

35°, To bisect a given spherical angle. 

36°. Ata given point in a given arc, to make a spherical angle 
equal to a given spherical angle. 

37°. To describe a circle through three given points upon the 
surface of a sphere. 

38°. To find the poles of a given circle. 

39°. Through two given points, A, B, and a third point, C, on 
the surface of a sphere, to describe two equal and parallel small 
circles ; the points A, B, C, not lying in the circumference of the 
same great circle. 

40°, To describe a triangle which shall be equal to a given spher- 
ical polygon, and shall have a side and adjacent angle the same 
with a given side and adjacent angle of the polygon. 

41°. Given two spherical arcs together less than a semicircum- 
ference, to place them so that, with a third not given, they may 
contain the greatest surface possible. 

42°, Through a given point to describe a great circle which shall 
touch two given equal and parallel small circles. 

43°. To inscribe a circle in a given spherical triangle. 


SECTION SECOND. 


Projections of the Sphere. 
PROPOSITION I. 


The Orthographic Projection of every circle of the (573) 
sphere, as meridians and parallels of latitude, will be an ellipse, 
circle, or straight line, according as its plane shall be oblique, 
parallel, or perpendicular to the plane of projection. 


1°, For let M, N, P, Q, be the segments of any 
two chords, M+ N, P-+ Q, of a circle, and m, 
n, p, q, their orthographic projections upon the 
plane of m+n, pt+q; that is, projections made 
by perpendiculars let fall from the extremities of 
M,N; P, Q, we have, 


M e N=P 2 Q, 
m = Mcos(M,m), n = Neos(M,m) ; (574) 
p= Pcos(P,p), ¢ = Qcos(P,p) ; : 
mn PY (575) 


cos*(M,m)  cos*(P,p) ’ 

which determines the nature of the curve of projection. 
2°, If M+ N pass through the centre, O, and ’ 

P+ Q be at right angles to M+ N, then P will Ge ke 
= Q, and... (574) p will = q; whence it follows hee 
that m-+n bisects a system of parallel chords {} 
p+q, and is itself bisected in 0, the projection of GD 
O; therefore 0 is the centre of the curve, As- Z— > 
suming o as the origin of a system of oblique co- 
ordinates, (575) becomes, 

(A+2z)(A-2z)_ yy 

cos*(M,m) os?(P,p)’ 
putting m-+2=2A; but R being the radius of the circle and B 
the projection of that R which is parallel to P, we find 
A = Reos(M,m), 
B+ Reos(P,p); 


ORTHOGRAPHIC PROJECTION. 28] 


R? R? 
es ae (4° — 2") = Fe °y; 
2 2 
or as ees |, (576) 


3°. If R be assumed in such position that its | 
projection, A, or, for the sake of distinction, a, Sata 
shall be parallel] to it and consequently 6, the new ree 
value of B, perpendicular to a, there results Ot AP op 


a =Rceos(M,m) = Reos0° = R, * 
b = Reos(P,p) = Reosl, eu 
I being the inclination of the circle to the plane of projection; we 
have also 
goo a 
o tae 
which is the equation of an ellipse of which the axes are 2a, 2h; 
and the proposition is demonstrated, since when J = 0, b becomes 
parallel to and equal to R =a, and when [= 90°, b becoming = 0, 
the ellipse vanishes in a straight line. 

Cor. 1. The ellipse may be referred to a system of ob- (579) 
lique codérdinates such that its equation (576) shall be of the same 
form with that obtained for rectangular axes (578), and 2A, 2B, 
mutually bisecting each other and all chords drawn parallel to them, 
are denominated Conjugate Diameters, of which the axes 2a, 20, 
are but particular values. 

Cor. 2. If two systems of parallel chords intersect each (580) 
other in an ellipse, the products of their segments will be propor- 
tional (575); and this property may be extended to the case in 
which the points of intersection lie without the curve. 

Cor. 3. An elliptical arc, MmApP, being given, the (581) 
centre, O, may be found by bisecting AOB drawn 
through the middle points of any parallel chords, 
MP, mp. N 


= iy (578) 


PZ = A 


Fig. 122%. 
Cor. 4. A conjugate, CD, to any diameter, AB, may (582) 
be found by drawing CD through the middle points of AB and PN 
a chord parallel to AB. 


282 ORTHOGRAPHIC PROJECTION. 


Cor. 5. A parallelogram, MPNQ, may be described in (583) 
an ellipse by drawing chords, MQ, PN; MP, QN, parallel to the 
conjugates AB, CD. 

Cor. 6. The diagonals MN, PQ, mutually bisecting each (584) 
other in O, are obviously diameters, and P may be so chosen that 
MN shall be the conjugate axis = 2a, in which case PM, PN, are 
said to be Supplementary Chords. 

Cor. 7%. The tangents drawn through the extremities (585) 
A, B, C, D, of the conjugate diameters, are parallel to the supple- 
mentary chords, and to the corresponding diameters ; and, by their 
intersections, constitute a parallelogram circumscribing the ellipse. 


PROPOSITION II. 


To make an Orthographic Projection of the Sphere. 

With the radius OA, equal to that of the re- 
quired map, describe the meridian AZP,H,B 
P.H,, passing through the north and south 
poles, P,, P,, and, consequently, perpendicular 
to the horizon seen edgewise in H,H.. Take 
the arc H,P, measuring the elevation of the 
pole, equal to the latitude of the place Z over 
which the observer is supposed to be situated at a very great [in- 
finite] distance. Having graduated the meridian and drawn the 
parallels of latitude ACB, which will be perpendicular to the axis 
P,OP., drop their extremities and centres, A, C, in the projections 


Fig. 123. 


a, c, also project the elevated pole P, in 2p, Hn 
Transfer the points a, c, p,, to the central me- ie 


ws 


ridian H,OH, of the map, and through cdrawa ”™ 
perpendicular to H,H,, intersecting the circum- 

ference in m, m.,; cm and ca will be the semima- 

jor and semiminor axes of the elliptical parallel H, 
of latitude mam,, which may be described by the Fig. 1232. 
first exercise under the ellipse. Next, for the 
meridians, let H,P,H,, = 4 be the angle which any 
required meridian, P,H, makes with the vertical 
meridian P,H,,, and H,H, = H, the corresponding 
arc intercepted on the horizon, further P,,H, will 
= 1, the latitude of the place upon whose horizon 
the projection is made; therefore, by Napier’s 
Rules, P,,H,,H, being a right angle, we have 


GNOMONIC PROJECTION. 233 


cosl,= coth cotH.,, 
or tan H = sinl tanh. (586) 


Having laid off H,H, = H found by calculation, Hn 
and drawn the diameter H,O, and Oz perpendic- 
ular to it, transfer the semimajor axis H,O to the 
straight edge of a thin ruler or slip of paper, and, 
having applied this line in p,z, and marked its 
point, y, of intersection with OH,, proceed to de- Hi 
scribe the meridian H,p, according to the first exer- Fig. 1249. 
cise under the ellipse.* | 


The work now described is to be combined in a 
single figure, when the features of the spherical 
surface will be laid down according to the projec- 
tions of the meridians and parallels of latitude. 


Fig. 125. 


PROPOSITION III. 


To make the Gnomonic Projection when the dial is horizontal. 

Let H,H,H, be the horizontal face of the dial, O its (fig. 124) 
centre and OP, the style elevated according to the latitude of the 
place ; the shadow at noon will fall upon the north point, H,, and 
its positions, OH,, for all other hours will be determined by (586), 
where / = the hour angle. 


PROPOSITION IV. 


To make the Gnomonic Projection of the Style upon a vertical 
south plane. 


The construction will be similar to the preceding, (fig. 124) 


observing that, 
tan(XII)T = cos/ tanh (587) 


* An instrument very convenient for this purpose may be constructed, consisting 
of a slender ruler carrying a pen or pencil in its extremity and furnished with two 
moveable pins, one to glide in a groove cut in the stem of a wooden T and the other 


along its top. 


284 STEREOGRAPHIC PROJECTION, 


PROPOSITION V. 


If an oblique circular cone be truncated by a plane at (588) 
right angles to that plane which passes through the axis and is 
arpan ater to the base, the section will, in general, be an el- 
lipse. 


Let AoOBy be the truncating plane perpendic- Vv 
ular to the plane, AMVBN, passing through the 
axis at right angles to the base; draw oy = y per- B 
pendicular to AB, and pass the plane, MNy, par- N 
allel to the base, it will be a circle, and the com- 
mon ordinate, oy =y, will be perpendicular to Fig. 126. 
the diameter, MoN ; 

(oy)(oy) = (oM)(oN). 

oM _ sinoAM oN sinoBN. 
oA sinoMA’* oB sinoNB’ 
, puting OA — OB = a, Go=z,-and b= yl}, 
sinoAM — sinoBN a 
~ sinoMA° sinoNB keicentt) s 


sinoAM , sinoBN 


But 


ee 


py ae By es 
I Mee ~sinoMA ° sinoNB é 
2 
_ = 1—-—,,, the equation of an ellipse. 


Cor. The section becomes a circle when the angles, (589) 
which the truncating plane and circular base make with the sides of 
the cone, are equal and contrary. 

For, when the angle oAM = oNB, then Z oBN = oMA, and the 
equation becomes 

yy? =a? — 2’, that of a circle, 


PROPOSITION VI. 


The Stereographic Projection of any circle of the (590) 
sphere is, in general, itself a circle. 

In the stereographic projection every point of the spherical sur- 
face is thrown upon an equatorial plane, denominated the primi- 
tive circle, by the intersection with this plane of a line drawn 
through the point and the eye situated in the pole of the primitive. 

Now, let AB be that diameter of any circle, 


STEREOGRAPHIC PROJECTION, 285 


whose extremities lie in the circumference, 
AP,BOH,P,EH,, passing through the poles, 
E, O, of the primitive H,H,H,H,, whose cen- 
tre, 0, is the same with the centre of the sphere, 
and draw AE, BE, intersecting H,H, in a, 6; 
ab is the stereographic projection of AB. 
We have, 


Fig. 127. 

measure of /EBA = tarcEH,A =+(EH,+ H,A)= meas. of Z Ead ; 
ZEBA = Ead, 

and (589) the projection of the circle described on AB is itself a 


circle situated in the primitive H,H,H,H,, and having the diame- 
ter, ab. 


PROPOSITION VII 


The distance from the centre of the primitive at which (591) 
any point will be projected, will be equal to the tangent of half 
the arc intercepted between the point and the pole of the primi- 
tive, the radius of the sphere being taken for unity. 

For let A be any point on the surface of the sphere ; we have 
oa _ sinoKa 
oF sinoaE 

or oa = tanzOA, [oE = 1}. (591) 

Cor. 1. If m be the centre of the circle, ab, we have, 

dist. of centre om =4(oa-+ 0b) =4(tantOA + tansOB), (592) 
where it is to be observed that oa, od, or their equivalents, tantOA, 
tantOB, change signs on A, B, passing O. ‘ 

Cor. 2. The radius ma = 4(tantOA — tandOB). (593) 

Cor. 3. The projection of a circle parallel to the primi- (594) 
tive will be concentric with the primitive. 

For, if A,B, be parallel to H,H,, we have 

OA,=OB,, .. om, =+4(tantOA, — tantOA,) = 0, 
Mod, = +(tantOA, + tantOA,) = tantOA,, 

Cor, 4. If the pole of the circle be in the primitive, the (595) 
distance of its projected centre will be the secant of its radius, and 
the radius of projection the tangent of the same arc. 

For, if we put H,A, = H,B, = wu, 
then 0A, = 45° + 4u, JOB, = 45° — 4u; 
and .%. (592), (593), reduced by (356), become 


= tanoEKa, 


286 STEREOGRAPHIC PROJECTION. 


PA: 
om, = —— = Secu, 
Ccosu 
sinu 
M343 = 
COSU 


== Ani, 


Cor. 5. A great circle perpendicular to the primitive (596) 
will be projected in a diameter of the primitive. 

For we have (595), 

OM, — 9 = Sec90° = infinity, 
MA, 9° = tan90° = infinity. 

The same is obvious from the figure. 

Cor. 6, Any great circle will have for the distance of its (597) 
projected centre, the tangent of the arc measuring its inclination to 
the primitive, and the radius of projection will be the secant of 
the same arc, 

Let P,P, be the diameter of any great circle, P,H,P,H,, and 
imitate the reasoning in (595.) 

Scholium J. Any point situated in the primitive, is its (598) 
own projection. 

Scholium I. Uf a great circle and the primitive intersect (599) 
each other in the diameter, H,H,, and a third circle passing through 
the axis of the primitive in the diameters, P,,P,, H,H,, the points, 
H,, H,, may be found by spherical trigonometry, and the point, p,, 
the projection of P,, having been determined by (591), there will 
be three points given through which to pass the circumference, 
H,p,H,, the projection of H,P,,H). 

H,,H, will be found (586), 
if we put HH) =, BLA 4,0, =. 


PROPOSITION VIII. 


To make the Stereographic Projection of the Sphere. 


I. Let the eye be situated at the south pole in 
order to project the northern hemisphere. De- 
scribe the primitive and graduate its circumfer- 
ence according to the number of meridians which 
it is intended to lay down; these will be diame- 
ters passing through the several points of division ‘ 

(596). | Fig. 128. 

For the parallels of latitude, we have only to count off from the 


STEREOGRAPHIC PROJECTION, 


287 


north point the corresponding degrees, and from the points of 
division to draw lines to the south point, and the intersections thus 
formed with the east and west diameter will be in the circumfer- 
ences of the circles of latitude whose centres will be that of the 


primitive, (591), (594). 


II. Let the eye be in the equator. 
primitive will pass through the poles, N, 8, 
and the central meridian, NS, and equator, 
EQ, will be north and south, and east and 
west lines intersecting in the centre of the 
Graduate the circumfer- 
ence as above, and, laying -the corner of a 
wooden square upon a point of division, di- 
rect the edge of one arm through the centre, 


primitive, (596). 


The 


Fig. 129. 


The intersection of the corresponding edge 
of the other arm with the production of the line, NS, will be the 
centre, and the distance of this point to the point of division the 


radius, for the description of a parallel of latitude, (595). 


The radii 


just employed set off from the centre upon EQ will give the centres 
for describing the meridians, which will pass through the points, 


N, S, (597), (598). 

III. When the eye is sit- 
uated otherwise than as 
above, consult (598), (599), 
and figure 130. 

Scholium. It will be ob- 
served that the middle of the 
map is comparatively con- 
tracted in the stereographic 
and enlarged in the ortho- 
graphic projection. To 
avoid this, the lines, NS, 
EQ, are sometimes divided 
into equal parts, and the 
map thus constructed is im- 


SO 
Were 


"Z 
IN 


Sy 


<F 
SS 


bk 
HEE? 
SHA 
Ss 
se 


Fig. 130. 


properly denominated a globular projection, 


PROPOSITION IX, 


To make the Conical Projection. 


288 . CON CAL PROJECTION. 


In this projection, the eye, situated at the centre of the sphere, 
throws every point of its surface upon that of a tangent or secant 
cone, the axis of which is coincident with the axis of the sphere. 

Let a sphere and cone be generated by 
the revolution of the arc, NnsES, and the 
straight line, ons, intersecting each other in 
n, Ss, about the common axis, vNo,o,OS, O 
being the centre of the sphere, N and S the 
poles, E a point of the equator, and no,, S0,, 
perpendiculars upon NS. 

1°. For the magnitude of a parallel of lati- 
tude, putting 

OE = Os = On = ON = OS = radius = 1, | 

arcA°=length of an are of A° in the Fig. ai 

parallel through n, 
arc A° =do. through s, arcA% = do. through E; 
l,, = latitude of n, 
l, = latitude of s, 


we have, 
no, =sinnN = cosi,, 
so, =sinsN = cosl, ; 
a) arc A®. + 4? = cosl,: cosl,, (600) 
and ercA®;: Ay = cosl, : 1, 


arc As..;. AS = cosl; : 1. 


2°, For the radii, on, vs, with which to de- 
scribe the sectoral surface, snNn,s.S8, the de- 
velopment of the conical surface upon a plane, 
we have, 


on : no, =sin90° : sinnvo,, 


or vn : cosl,=1 : sint(sS—nN); 
Fig. oe 
cf radius vn = pelieohhet. 
sind(/,+1,)’ 
and radius vs = ck | ail 
sind(d,+ | aids) ar 


3°. To find the number of degrees in the angle, svs., corre- 
sponding to a difference of latitude of A° ; with the trigonometrical 
radius, vr = 1, describing the arc, rr, there results, 


CONICAL PROJECTION, 289 


arcrr, : radius vr = arcnn, : radius vn, 


ead me cosh, os 
or are? °F 1‘ are AP’: sinthslady . 
arcv° =arcA;, « sind, + /,), (602) 


Now, to make the plane of projection approximate the spheri- 
cal surface as much as may be conveniently done, take nm and 
s so that these and the middle point of the map shall divide its 
meridian equally, or as nearly so as the degrees of latitude or their 
aliquot parts will permit. Calculate the radii, vn, vs (601), and 
find the degrees, v° (602), corresponding to the degrees of longi- 
tude, A°, to be embraced by the map; from a scale of the required 
magnitude, lay off the meridian, NS = vs — vn, and produce it in 2, 
about which as a centre, and with the radii, vn, vs, taken from the 
same scale, describe the arcs, nNn., sSs,, and, having made the arc, 
rr, = v°, and drawn vns, Vilgs., divide the arc, nNn,, into A parts ; 
through these points of division and corresponding ones in NS and 
NS produced, draw the remaining meridians and parallels of lati- 
tude. 

4° To find the half chords, cn, cs, and their altitudes, Nc, Sc. 


ne: nv=singv : sin90°, 


: cosl, 
half chord cn = sindv ® sind(1, +1.) 9 (603) 
and 6s : cn = cosl, : cost,. 
Again, vc : von =costv : 1; 
vc = COSstv « ennai 
: ind(l, + 1,)’ 
ssibytie eagyeeyind sodtar a 
Ne = (1 — cos}v) siatiliecaia’ 
ened, 2 cosl,, 
or Ne = 2sin*1v Fibs rr 
wig wel 4 cos, 
so Sc = 2sin?4v andl olt iy’ 
Finally, he pe a A vp ae (605) 


sind(d, + l,) 

Forms (603), (604), (605), enable us to make the construction 
with accuracy when the point, v, is so distant as not to admit of 
employing the radius, vn, by finding the successive values of v° 
(602) for A = 1°, 2°, 3°, 

Scholium I, This Peaiection is well adapted to the construction 

19 


290 EXERCISES, 


of maps of moderate extent from north to south; and when a 
greater number of degrees of latitude are to be embraced, it will 
also become quite accurate by regarding the conical surface as 
composed of several placed end to end, joining each other in paral- 
lels little distant, as 1°, 2°, or 30. 

Scholium II. When the points, n, s, are situated at equal dis- 


tances on opposite sides of the equator, E, the cone becomes a 
cylinder ; for we have, 
J cosl 

l.=—~l, «. (601), radius p2= 


~ = infinity, 
and (605), NS = 2sinl,, 


sinO° 


Scholium UI, When the points, n, s, are taken on opposite 
sides of the pole, N, and at equal and moderate distances, the pro- 
jection becomes a circular plane, well calculated to represent the 
circumpolar regions, either of the earth or the heavens. 


EXERCISES. 


1°, Construct a map of the sphere as it would appear to an ob- 
server elevated to the zenith of lat. 43° N., lon. 78° W. 

2°. Project the sphere as, supposing it transparent, its opposite 
surface would appear on the rational horizon to an eye situated in 
the surface of the place just given. 

3°. Make aconical projection of the State of New York and 
the countries extending a degree or two on the north and south, 
assuming the centre of the map near Syracuse, and calculating the 
radii to each degree of latitude. 

4°, Prove that the section made by a plane through the vertex 
of a right circular cone, is an isosceles triangle, passing into a 
straight line by a revolution of the plane, 

5°, Show that if a plane cut a right circular cone parallel to its 
side, the section will be a parabola, vanishing in a straight line. 

6°. The truncation of a right circular cone will be found to be 
an ellipse, passing on the one side into a circle and a point, and on 
the other into a parabola. 

7°, If a right circular cone be cut by a plane not passing through 
the vertex, not truncating it, nor parallel to its side, the section 
will be a hyperbola, whose limits will be a straight line or para- 
bola. 


BOOK THIRD. 


NAVIGATION. 


SECTION FIRST. 


Problem of the Course. 


The earth is an oblate spheroid, formed by the revolution of an 
ellipse about its minor axis, the polar diameter being 7899'170 and 
the equatorial 7925‘648 English miles ; but, for the purposes of 
Navigation, it may be regarded as a sphere with a radius of 343775 
geographical miles or minutes, 

If a ship could be conveniently guided in the arc of a great circle, 
which would be the most direct path, nothing would be easier than, 
by Spherical Trigonometry, to determine her course from one 
point on the earth’s surface to another, or, having given the point 
of departure and the distance run, to find her place. But as, in 
practice, the vessel is directed by the magnetic needle so as to cut 
all the meridians at a constant angle, the path actually described is 
acurve of double curvature, denominated the Rhumb Line, or 
Loxrodromic Curve. 


PROPOSITION I 


To find the difference of latitude. 


Let a ship sail from X, to X and describe the 
rhumb z, intersecting the meridians z,, 7, 7 -+h, 
at the constant angle c= course. P is the pole, 
x, the polar distance of X,, 2 that of X, and y 
= / (x,,x) = the difference of longitude ; &, h, ¢, 
are the vanishing increments of y, x, z, and 7 is 
a corresponding vanishing are described about 


Fig. 132. 


292 DIFFERENCE OF LATITUDE AND LONGITUDE. 


P with the radius, x, which may therefore be regarded as the are 
of a great circle perpendicular to r+ Ah. 

It is required to find the function x = ¢z. 

Assuming the earth’s radius for unity, we have 


sin(90° — c) 
L a= be = [+] = COSc 


~ sin90° | 
ae x = zcosc + constant, 
and .*. 2,5 Bid = consiant | 
T— 2, =.2 cose, (606) 


or Difference of Latitude = Distance < cosine of Course. 


PROPOSITION II. 
To find the difference of Longitude. 


Let y = yz be the function required; there results, 


yoo CE-U8)-FEI-C) EH 


A sin90° sinc 
~ sing — sin(90°—c)’ 
} 1 sinz sint 
or Y ya te = tance pare = tanc e (sina)? = imeoss e tance, 
1+ cosz ned eh mei(Bose pratt F sina ter 
put [ECHR ) [(; na es) ~ 1—cos?z 1 — cos?z’ 


y= [2 (= a ey MW « tanc-+ constant, 


1 — cosz 
1+ cosr\-}_ (1 —cosz\4 — /(1— cosr)?\4 
Bits al ki — om) ve) (; + a) it ( 1— | 
eee sala = fandz; 
sing 


y = tanc « / tant + constant, 


ands", 0 = tanc « 7 tantz, + constant ; 
tantz 
y = tance « ] —— 
tandz, 


But, in order to adapt this form to the purposes of computation, 
we must convert the Napierian into common logarithms, and adopt, 
as is customary, for the unit of measure the nautical mile or minute 
instead of the earth’s radius, which will be done by multiplying by 


i = 2'30258509 and 3437‘75; and there results, 


MERIDIONAL PARTS. 293 


tantz 


dif. lon. y = 7915°705 tanc « log. i (607) 
[log. 7915‘705 = 38984896]. 
Cor. 1, If X, be a point of the equator, or x, = 90°, 
then, y = 7915*705 tanc log. tan(45° + 4), (608) 


putting latitude (x — 90°) =1., 

Cor. 2. If we imagine the 4 
spherical surface rolled out upon | 
a plane so that the equator, y, .- 
shall become a straight line, the 
parallels of latitude and their dis-_ . 
tances apart being increased for ‘ 
the purpose, the distance, z,, the 
difference of longitude, y, reckon 
ed on the equator, and the increas- 
ed difference of latitude, m, esti- 
mated from the equator, will form 
the three sides of a right angled triangle, and we find 


Fig. 133. 


tanc=~, or y = m tance ; 
m 
*, (608), =m ='7915°705 log. tan(45° + 41). (609) 


When a table is constructed by aid of (609), called Meridional 
Parts, the problems in navigation relating to the course may be 
solved by the rules of plane trigonometry ; and a Chart constructed 
from these numbers, bears the name of Mercator. 


PROPOSITION III. 


To find the difference of longitude when the ship sails on a 
parallel of latitude. 

In this case cosc becoming = 0, form (606) will be inapplicable ; 
but since z is now a parallel of latitude and x = z,, = a constant, 7 
coinciding with 7, we find, 


[i |- [- hy sin90° _ tl 3 
Wan = is sinz,  sinz’ 


I~ sing, * 
which accords with (600). 


(610) 


294 EXERCISES. 


Scholium. There are various methods of solving the Problem 
of the Course. 

1°. Plane Sailing, where the distance run being small, the sea 
is regarded as a plane, and the common forms of plane trigonome- 
try are employed, 

2°. Traverse Sailing, where several short courses are run in 
succession, and forms (421), (422), are applicable, 

3°. Parallel Sailing, where the ship sails on a parallel of lati- 
tude, and the difference of longitude will be found by (610). 

4°, Middle Latitude Sailing, where the distance run is mode- 
rate and nearly east and west; in this method the middle latitude 
is assumed to be equal to the half sum of the two extreme lati- 
tudes when both are of the same name. 


5°. Mercator’s Sailing, as explained in (609), 


EXERCISES, 


1°. A ship sails from 3° S, lat. and 9° W. Ion., N. 40° W. 538 
miles. Required her latitude and longitude. 


For the difference of latitude we have (606), 
log. cos 40° = 1884254 
log. 538 = 2°730782 
dif. lat, = 41213’, 2615036 
For the difference of longitude (607), 
389849 


log. tan 40° = 1+92381 
log, tantz =002950 -gouth 


log. tanta, = 197725 L Pole. 
tantz 


log. = 05225 log. lo OL | = 2°71809 
tan4z, 
254039 


dif. long. = 347 miles. 
2°. In what direction must a ship sail from New sei to arrive 
at London, and what distance will she make? 
3°. Aship sails from 20° N. L. and 70° W. L. for the Azores 
in 38° N. L. and 30° W. L. After performing a thousand miles 
of her course, she tacks for New York; in what direction must 


PROBLEM OF THE PLACE, 295 


she steer, and what will be her distance to the last-mentioned 
port? ; 

4°, Ifa ship could sail from the equator N. 45° E. till she should 
arrive at the pole, what would be her distance run, and what her 
progress in longitude ? 

The distance run will be found = '7636‘68 miles, a finite num- 
ber, but the difference of longitude will be infinite; that is, the 
ship will make an infinite number of turns about the pole, spinning 
there at length, like a top, with an infinite velocity of rotation, 

5°. Calculate the following table of meridional parts, and em- 
ploy it in solving the exercises just given, and in constructing a 
chart extending from the equator 60° each way. 


Table of Meridional Parts. 


|329%2 | 


3203 | 


4183 
4294 


10° | 20° 30° | 40° 60° | 70° | s0° | 
603 1225 [1888 '2623 13474 14527 5966 8375 
664 |1289 |1958 |2702 3569 4649 6146 8739 
725 |1354 [2028 2782 3665 |4775 |6335 9145. 
787 |1419 |2100 |2863 3764 |4905 6534 9606 
848 |1484 (2171 [2946 3865 | 

3030 3968 | 

13116 4074 


4409 


SECTION SECOND. 
Problem of the Place. 


Owing to the impossibility of sailing absolutely in the required 
course and measuring accurately the distance run, it becomes neces- 
sary to resort to astronomical observations for the purpose of de- 
termining from time to time the ship’s position on the ocean, To 
this end the daily places of the most conspicuous of the heavenly 


296 LATITUDE. 


bodies are computed and laid down in the Nautical Almanac* for 
Greenwich mean time, and for that of Paris in the Connaissances 
des Tems.t These places are such as they would appear to an ob- 
server at the earth’s centre, and are estimated in right ascension and 
declination, 

The declination of a heavenly body is its angular distance from 
the Eguinoctial [Celestial Equator], plus if north, minus when 
south, and the distance measured, commonly in hours, from the 
Vernal Equinoz, to the east upon the Egquinoctial, is the right 
ascension of the same body. 


PROPOSITION I. 


Given the geocentrec meridian altitude of a heavenly body and 
its declination, to find the latitude. 


Let O be the centre of the earth, P the pole ele- 
vated above the horizon, H, and 90° distant from FP Ss 
, ; : S 

the celestial equator, E; then denoting the zenith gy. 
by Z and the position of any heavenly body in the Fig. 134. 
same meridian by 8, S, S, we have 

ZOE = HOP = HOS = POS, or, 

ER Fe EIS RTA 

The latitude of a place, or the altitude of the elevated (611) 

pole, is equal to a heavenly body’s meridian distance from the 
point of the horizon under the pole, increased or diminished by 
the polar distance af the same body, according as it is situated 
below or above the pole. 


Scholium I. The polar distance of any heavenly body is (612) 
equal to 90° = its declination. 

Scholium ITI. The latitude may be found on shore, or when (613) 
the vessel is stationary, by observing that the elevation of the pole 


* Blunt’s reprint is sufficient for all purposes except those of a fixed and regularly 
furnished observatory, and is much less expensive. The student should be possessed 
of such astronomical knowledge and acquaintance with the stars as may be acquired 
from elementary works like “‘ Kendail’s Uranography.” 

+ To make use of either ephemeris it will evidently be necessary to reduce the time 
of any observation to that of the meridian for which the ephemeris is calculated, by 
allowing 15° to the hour, and observing the equation of time. 


LATITUDE, 297 


= the half sum of the upper and lower meridian altitudes of any 
circumpolar star; since 

P,,=(S,),— 8S, and P, = (S,),+ 8, 
give P=4H((S,).+(S,)- 

Example. On the 20th of August, 1847, at noon, the meridian 
altitude of the sun’s centre was found to be 62° 49’ 50”; at the 
same instant the chronometer, regulated to Greenwich time indica- 
ted 2h. 45m. Required the latitude of the place. 

Interpolating (279) for the declination from the Nautical Almanac, 
we have the following 


Operation. 
Day of |; Declination. 
Month. i ' : 
20 yy = 12 35 190 19 492=D, 
21 12 15 2958 20 008 +116 = D, 
22 11 55 290 114 .[.. Ds = 90] 
23 11.35 168 79 12*2 
21 45 
xg = 2h, 45m. = nh = 41d, 
Di 0. Age 1D, =-+ 63 
11 19 492 53 
96)209 541°2 96)58'3 [+ 44 
zD,=—2 169 ‘6 


~% [-Gt-1)] 


[e) / 4 
yo = 12 35 1940 


zD,= —2 1693 
D. 
a(x —1)° en — 5 


N. D.=y, = 1233 22 
N. P. Dist. = 77 26 57'S, [—], (612) 


——————_ 


S’s alt. = 62 49 50°0 


S,= 117 10 100 


39° 43’ 12°2" Lat. N., required. 


298 TIME, 


PROPOSITION II 


Given the polar and zenith distances of a heavenly body, and 
the zenith distance of the pole [= colatitude], to find the time. 


The hour, found approximately by the longitude of C 

the vessel and the chronometer, regulated to Greenwich at 

time, is to be corrected by observation. nine 
Let A be the pole, B the position of the heavenly 

body, and C the zenith of the place ; then the hour an- 

gle A will be found by (540) or by (541), the first being Fig. 135. 

preferable when A is small, (614) 
Example. In the preceding example, after sailing 3 hours due 

west, the zenith distance of the sun’s centre was found to be 

47° 43' 51:3". Required the hour angle, A, or the time referred 

to the meridian of the place arrived at. 


Operation. 
3h.=dd. 8) — 19 49:2 8) +153 
2 28'65 ‘66 
8 
Pe de tg] 
re) i / 
12 33 282 
— 2 28°65 
— 58 


N.D. =12 30°33% 


a =47 43 51 


b= 50 16 478 
c= 717 29 270 


2)175 30 61 
h=87 45 305 


——- 


TIME. 299 


12) / i — 
h—b=37 28 15:25 1°78 41 595 


h—c=10 15 36:05 1:25 07 018 
b=50 16 478 188 60 257 
c= 77 29 270 198 95 661 


2)115 92 695 


B rie oy 157 96 347 
4 $A = 22 19 32:2 
and A= 44 39 44 
or A = 2h. 58m, 3653s., time required. 


Therefore the ship had sailed 20‘922 miles. 


Scholium I, The time of setting of a heavenly body, or, rather, 
the hour angle when the body is 90° distant from the zenith, will 
be better found by (569), since we have 


cosA, = cotd cote, 
or cos(180 — A) = tan lat. x tan dec., (615) 


where it is to be remembered that the declination will become 
minus when measured in a direction opposite to the elevated pole. 
Ex, When did the sun set to the place in the last example but 
one? Ans. 6h, 42m. 38s. 
— 19s. for change of dec. 
Scholium II, The time of the apparent setting of aheav- (616) 
enly body will be found by substituting for a, the apparent zenith 
distance of the horizon, which will generally differ from 90°, 
Scholium III, The hour when twilight ends will be found (617) 
by putting a = 90° + 18° = 108°, since the sun is 18° below the 
horizon at that time. 
Ex. When did twilight end in the above example ? 
Ans. Th, 3m, 27s, 
Scholium IV. The azimuth, C, or bearing of aheavenly (618) 
body, when its zenith distance, polar distance, and either the hour 
angle or the zenith distance of the pole, are known, will be found 
by (551), or by (540), 
Ex, What was the bearing of the sun at setting in the above ? 
Ans. 73° 40’ 14”, 
How might the variation of the magnetic needle be determined ? 
Scholium V. When on shore, the magnetic variation will be de- 
termined, with greater accuracy, by ascertaining the greatest east- 


300 VARIATION OF MAGNETIC NEEDLE, 


ern or western elongation of a circumpolar star, 7. e., its azimuth 
when B= 90°. We have (568) 
cosc, = sind sin C, or log. sinC = log. sinc —log. sind. (619) 

For this purpose the pole star is preferable. Its polar distance 
on the Ist of January, 1847, was 1° 30’ 22‘85", and decreases 
19‘273" annually. 

The time of greatest elongation will be found by taking the dif- 
ference of the right ascensions of the sun and star from the Nauti- 
cal Almanac. The elongation will be easily observed by the aid 
of any altitude and azimuth instrument, carefully levelled, as a 
theodolite—or, if no such instrument be at hand, by suspending 
a long plumb line, with its weight swimming in a pail of water to 
prevent agitation by the wind, and, at a suitable distance south of 
the line, placing upon a table, a piece of board carrying a sight 
vane, by the motion of which, east or west, the required point will 
be obtained when the star appears no longer to depart from the line. 
Next alline a lighted candle at a considerable distance north—ten 
or a dozen rods—-which is to be blown out and left till morning, 
when it may be observed with a common compass, or the true me- 
ridian may be traced by calculating a triangle, and constructing it 
with a rod graduated for measuring lengths. In making the above 
observation it will be necessary to illumine the plumb-line in the 
direction of the star, 

Scholium VI. When the zenith distance, polar distance, and 
either the hour angle or the azimuth, of a heavenly body are given, 
the latitude may be found by combining (535) with (383 ... 4); for 
we obtain 

cose COS COSZ 


tang = + , and sin(O + ¢) +e 620 
sinc cosA’ Dt 2} sinc cosA’” hale 
cosa cosc cosz 
or tang See andain(b + 2.) =|». 621 
* sina cosC’ Coed sina cosC (621) 


Having determined by the aid of a theodolite, on the same or 
different evenings, the altitudes and azimuths of the same or differ- 
ent stars, the mean of several computations by (621) will give the 
latitude of a place on land with a considerable degree of accuracy. 
The same observations will also make known the hour 

The error that is likely to be committed in the hour angle, A, 
the chronometer not indicating the exact time, may be corrected 
by trial, from a couple of observations, the first made soon after 


DOUBLE ALTITUDES, 301 


the meridian passage and the other several hours later. It is to be 
observed that a change of sign in the error of A will correspond to 
a change of sign in the error of 6, and that A is to be varied till 
two values of 5 are found answering to the difference of latitude 
made by the vessel between the observations. 

Scholium VII, Tux Meruop or Dousie AttirupEs has for 
its object to determine the latitude and time from two zenith dis- 
tances, either of the same body taken a few hours apart, or of dif- 
ferent bodies taken at the same time. 

Let CB be the first zenith distance of the sun, 
and C, the zenith arrived at when the second C,B, 
is taken; the number of miles the ship sails 
gives the number of minutes in the arc CC,, and 
the angle, BCC, = C, is determined by the course 
and the azimuth of B at the first observation, Fig, 1352. 

Therefore, in the triangle, BCC.,, to find c, we have 


cosc = cosb cosc, + sinb sinc, cosC ; 


but, as d is small and c differs little from c,, putting c = c, — d, and 
expanding by (317), reserving only the first and second powers of 
d and b, the reduced equation becomes, 

d? — 2tanc, « d= b? — 2d tanc, cosC, (622) 
or, better, C,—c=d=becosC, (623) 
when, as in the present case, the second powers of the small quan- 
tities, d, b, may be rejected. 

There are now three steps in the operation: (624) 

1°. In the triangle, BAB,, we have the two sides, AB, AB., and 
the included angle A = the time elapsed between the observations, 
or to the difference of right ascensions, if two stars are observed 
at the same time, to find the remaining parts ; 

2°. Therefore the angle, C,B,B, becomes known from the three 
sides of the triangle, C,B,B ; 

3°, And, lastly, in the triangle AB,C,, we have the two sides, 
B.A, B,C,, and the included angle, B,, to find the colatitude, AC,, 
and the hour angle, C,AB,. 

Note 1. When the declination changes but little between (625) 
the observations, the first part of the operation may be shortened 
by regarding the triangle, BAB., as isosceles, taking instead of AB 
or AB,, +(AB-+ AB.) for the equal side and solving by (570). 

Note 2. In any of the preceding operations, when a star, (626) 


302 LONGITUDE. 


instead of the sun, is employed, the hour, angle will be converted 
into degrees by observing that a sidereal day is equal to 23h. 56m. 
4:0906s. ; whence it follows that 


1 solar hour = 1‘002738 sidereal h. = 15°*04107, 


Note 3. Before taking out any quantity from the Nautical (627) 
Almanac, or the Connaissances des Tems, the time of the place 
where the observation was made must be reduced to the meridian 
and denomination of time specified in the ephemeris. ‘ Equations’’ 
(as these numbers are technically called) for reducing apparent to 
mean time, will be found in the ephemeris itself. 

Example. On the first day of August, 1847, at 10 o’clock 43m. 
25s. P. M., apparent time, and in 20° W lon., it is required to find 
the moon’s declination. 


Ap. time of pls. Diff. lon. Ap. time at Gr. 
10h, 43m. 25s, + 2£h. = 11h. 3m. 25s. 
Ap. time at Gr. Kq. time. Mean time at Gr. 


11h. 3m, 25s. + 6m, 1 s, =11h. 9m, 26s, 
6° 27 136" + 1’ 40'8" = 6° 28' 54:4” N.Dec. Ans. 
Example. At the time specified in the example preceding the 
last, the ship tacks and sails N 43° W, nine and a half miles an 
hour for two hours, when the zenith distance of the sun’s centre is 
found to be 37° 55’ 40”. Required the latitude of the place arrived 
at and the exact time. 


PROPOSITION IIl. 


The Lunar Method for the Longitude consists in find- (628) 
ing by the ephemeris the time corresponding to the geocentric 
distance of the moon from a heavenly body. The problem re- 
guires for its solution the apparent angle which these bodies make 
with each other, as well as their apparent and geocentric zenith 
distances. 


Let A, B, C, be the places of the zenith, moon, ! 
and some other heavenly body, as the sun, as seen ip 
from the earth’s centre. On account of parallar aon i 
and refraction, to be explained hereafter, the moon Fig. 136. 

will appear at B,, below B and in the same vertical, and the second 
body will be seen in C,, vertically above C; hence B,C, will be the 
apparent distance of the two luminaries, 


LONGITUDE. 303 


There are now three steps in the operation for the lon- (629) 
gitude : 

1°, In the triangle B,AC, we have the three sides dy, b., c,, the 
apparent distances of the two luminaries and the zenith, to find 
their difference of azimuth, A). 

2°. In the triangle BAC, we have the two sides, b, c, and the 
included angle <A, to find the third side, a, or the geocentric dis- 
tance of the moon from the heavenly body. 

3°. Lastly, with a, we enter the table of Lunar Distances and 
take out, by interpolation, the corresponding Greenwich time, the 
difference between which and that of the place, converted into de- 
grees, will be the longitude required. 

Scholium. The lunar distance may be calculated from (630) 
the right ascensions and declinations of the two heavenly bodies ; 
since the codeclinations and the difference of right ascensions con- 
stitute two sides and the included angle of a spherical triangle of 
which the third side is the distance required. 


Example. On the 9th April, 1837, at 5h. 29m. 36‘8s, mean time, 
the apparent distance of the centres of the sun and moon was found 
to be B.C, = a, = 55° 1’ 188"; at the same instant the apparent 
zenith distance of the moon was AB, = c, = 28° 49’ 50” the true 
AB = ¢ = 28° 27 48", and the corresponding data for the sun were 
AC, = 6, = 68° 9' 46”, AC = b = 68° 12’ 10”... The moon was east 
of the sun and both west of the meridian; the estimated latitude 
was 41° 47 N, and the longitude 2h. 10m. west, which gives, ap- 
proximately, 7h. 40m. for the Paris time. 


True distance BC = a = 55° 17’ 20'[ calculated]. 
We find in the Conn. des Tems, that 
at 6h. Paris time a=64° Tl 36"; 


difference 1° 5 44’; 
the change in a for 3h. was 1° 25° 55", 
1° 25) 65"; 1° 6 44° :.: Sh,.: Zh..17m, 42'9s., 
the corresponding hour at Paris was 8h. 17m. 42‘9s., 
and the exact hour of the place was 5h. 29m. 36‘8s. ; 


a 


-, the required longitude was Zh, 48m, 61s, W. 


N. B. This result may be corrected by commencing the calcu- 
lation anew with 2h. 48m, instead of 2h. 10m. 


SECTION THIRD. 


Description and Use of Instruments. 


Besides the Chronometer already alluded to, the principal instru- 
ments of the navigator are the Compass, the Log, and the Sez- 
tant, 


PROPOSITION I. 


To steer a determinate course. 


This is effected by aid 
of the Mariner’s Compass, 
which consists, essentially, 
of a circular card, poised 
ona pivot and carrying a 
magnetic needle, The cir- 
cumference of the card is 
divided into thirty-two equal 
parts, denominated points, 
and each point into quar- 
ters. ‘The bearings, called 
rhumbs,are estimated from 
the north and south toward Fig. 136.. 
the east and west, as follows ; } 

N, N by E, NNE, NE by N, NE, NE by E, ENE, E by N, E:; 
N,N by W, NNW, NW by N, NW, NW by W, NNW,W by N, W; 
S,S by E, SSE, SE by S, SE, SE by E, ESE, E by 5S, E; 
S, S by W, SSW, SW by 8S, SW, SW by W, WSW, W byS, W. 

It is easy, therefore, to convert any rumb into degrees ; 


thus, SE by E=S 5 points E=S 4- 90° E=S 56° 15’ E. 


PROPOSITION II. 


To determine the Ship’s Rate. 


SEXTANT, 305 


This is accom- 
plished by a line, 
divided at one ex- = 
tremity into two 
branches which Fig. 1360. 
are attached to the vertical radius of a wooden sector, having its are 
so loaded with a strip of lead as to swim buried in the sea with its 
face perpendicular to the line of draft. The instrument is denom- | 
inated the Log and Line. The log-line is divided into equal parts, 
called knots, of which 120 = a nautical mile; hence the number of 
knots run off from a light reel while a half minute glass is emptying, 
corresponds to the miles which the ship sails in an hour. 


PROPOSITION III. 


To determine the zenith distance of a heavenly body. 


I. The apparent altitude above the apparent horizon is observed 
with the sextant. This instrument, so use- 
ful to the surveyor and astronomer and 
indispensable to the navigator since it may 
be held in the hand without any fixed sup- 
port, consists essentially of an are embra- 
cing 60° and graduated as 120°, of an index, 
I, carrying upon the centre, O, a mirror, 
MON, of a second glass, PGQ, one half 
silvered, and of a telescope, T. Besides, 
there are screws for fastening and tilting the 
mirror, MN, for tilting and turning G, for 
tilting and elevating or depressing the tel- 
escope, I’, and for fastening the index, I, 
and moving it over a small arc when once 
clamped. 

Now let the index, I, take up such a po- 
sition that the moveable mirror, MN, shall 
be parallel to the fixed, PQ, and let a ray 
of light, HO, fall upon MN in a direction 
to be reflected in OG and again from G through the telescope, T, 
to the eye. Since, by a principle in Optics, the incident and re- 
flected rays make equal angles with the mirrors, we have 


20 


306 SEXTANT. 


Z HOM =GON 
and OGP = TGQ; ; 
/ HOM = TGQ = GTM; 


hence, the rays HO and HGT being parallel, the two images of 
the distant object from which they proceed, the one seen by reflec- 
tion from G and the other through the clear part, will appear to 
coincide in the direction TGH. But aray, SO, falling in any other 
direction, being reflected in OS,, will not enter the telescope, and, 
in order to make it do so, we advance the index, I, till arriving at 
7, OS, coincides with OG. There results, 
ZnOG =m0S = mOM+ MOS, 
nOS, = nON — NOS, = mOM — MOS; 
, GOS, = 2m0M, 
but HOS = HOM — MOS = GON — NOS, = GOS, ; 
Z HOS =2m0M = arc Ii, i. e., 
The angle measured by bringing into coincidence the (631) 
images of two objects, one seen by reflection and the other directly, 
is double the arc passed over by the index. 


The reasons for the following adjustments will now be (632) 
obvious. | 

First Adjustment. 'To make the index glass, O, perpendicular to 
the plane of the instrument. Bring the index to the middle of the 
graduated arc, and, looking into the mirror, observe whether the 
limb seen directly and by reflection appear broken or continuous. 

Second Adjustment. To make the horizon glass, G, perpen- 
dicular to the plane of the instrument. Having interposed the 
colored glasses, direct the telescope to the sun, or, better at night, 
to a star of the first magnitude, when, on moving the index back- 
ward and forward, the two images, in passing each other, should 
exactly coincide. 

Third Adjustment. To make the line of collimation, or the axis 
of the telescope, parallel to the plane of the instrument, Turn the 
- eye-piece till the two wires stretched parallel to each other in the 
common focus of the lenses for the purpose of restricting the ob- 
server to the centre of the field of view, become parallel to the 
plane of the instrument; next, bring two objects, as the edges of 
the sun and moon, distant by 90° or more, into coincidence upon 
one of the wires; the images should continue to coincide when, by 
turning the instrument a little, they are made to appear on the sec- 


ARTIFICIAL HORIZON. 307 


ond wire. The above adjustments should be repeated one after 
the other till all are quite accurate. 

Fourth Adjustment, To determine the index error, or to make 
the zeros of the vernier and limb agree. The error will be deter- 
mined by bringing the reflected and direct images of a star or 
planet into coincidence, or, by observing the readings when the 
images of the sun are brought into contact on opposite sides. 

Fifth Adjustment, To. make the reflected and direct images 
equally bright. Turn the wp and down piece which carries the 
telescope to and from the plane of the instrument, till the object is 
accomplished, 

For a description of the Reflecting Circle, see Franceur’s 
*“* Géodésie.”’ 

Now, for example, suppose it be required to find the ap- (633) 
parent altitude of the sun’s lower limb above the apparent horizon, 
Turn the shades which are situated in front of the mirrors into the 
lines OG, HG, so as properly to soften both the reflected and di- 
rect rays; next, set the index at zero and direct the telescope to 
the sun; push the index a little till the lower image is out of sight, 
following the upper one; drop the screenin front of the horizon 
_ glass that the sky may appear, and continue the motion of the 
index till the sun’s image touches the horizon ; clamp, and balance 
the instrument gently about the line TGH, alternately producing 
and breaking contact, the better to see when the tangency becomes 
exact, and this will be accomplished by carefully turning the tan- 
gent screw, which impresses a small motion upon the index. A 
microscope may be employed for reading the angle, and one is 
sometimes attached to the index. 

Scholium I. On shore an artificial horizon may be em- (634 
ployed, consisting of a basin of mercury, from the sur- g 
face of which the light of the celestial body is reflected. y \~~7 
The measured distance, SMS,, of the two images will ea 
be double the apparent altitude, SMH. Fig. 138. 

Scholium IT. As a convenience in finding the time, we (635) 
may choose a point of observation, and determine, once for all, the 
hour angle of a terrestrial object, sufficiently remote and distinct, 
from which to measure the angular distance of the sun, whenever 
required, 

Scholium III, In obtaining the data for the longitude, (636) 
three arcs, measured at the same instant, are required; viz., the 


308 DEPRESSION. 


altitudes of the moon and a star or the sun and the lunar distance. 
The three measures may, however, be executed by a single observ- 
er, by taking at equal and short intervals of time, 1°, the altitude 
of the sun or star; 2°, the altitude of the moon; 3°, the lunar 
distance ; 4°, a second altitude of the sun; 5°, a second altitude 
of the moon—from which the altitudes simultaneous with the lunar 
distance will be readily found. 

Il. For the depression of the horizon. Let A be the 
position of the observer, elevated above the level of 
the sea by the altitude h, let T be the apparent hori- 
zon, O the earth’s centre and r its radius, also let 
Z AOT =e, and the depression TAH [= AOT] = d. 
We find 


Fig. 139. 


sine AT JV(Qrh+ des 
cose OT rn 
which may be reduced to 


br atin Pid poarch tov igen O82 


observing that h is small compared Gritk T; ant, consequently, 


tand = 


tand = d nearly, also that an arc of 1”, 


648000 .. 
T 


ee T : tained 
= 180 - 60+ 60 1s Containe 


in the radius (= 1), times 


Now, from the aggregate of many observations made on shore, 
with an instrument adjusted by a spirit level, there results, 
Calculated. 


2 
log. ( “ ei re 
Calculated. 
24396000 feet. 177128 
24395000 « 1°77128 
24395000 « | 1°77128. | 


648000 


if 


Observed. Observed. 


10 feet. | 186°76" 
20 * | 264512 
30 <“ | 32348 


For calculating the depression then we have 
log. d' = 1‘77128 + dlog. h. (638) 
Scholium. It will be observed that the value 24395000 (639) 
is greater than the true radius, which, according to Dr. Bowditch, 
may be taken at 20911790 feet. We infer, therefore, that a ray of 
light, passing tangent to the sea, is bent into a curve concave to- 
ward the earth’s centre ; since (637), if the true value of r be taken, 


REFRACTION, 309 


h will be smaller than the value assumed in the figure, and more 
and more so the greater d becomes, / varying as the square of d; 
consequently the path of the ray will fall below the tangent, TA, 
and continually depart from it. 

Ill. For the refraction. We will suppose an observer provided 
with an astronomical clock, carefully regulated to sidereal time 
(24h. = 360°), and an altitude and azimuth instrument, accurately 
leveled so that the upper and lower meridian transits of a circum- 
polar star shall take place at intervals of twelve hours as exactly as 
possible, and small, unavoidable errors allowed for, The data, 
both observed and calculated,* requisite for determining the law of 
atmospheric refraction, may be tabulated as follows, 


[Last of Aug., 1847. ] 
Ap. zen. dist. of polaris, at upper transit = 45° 33’ 50”, 
do. do. do. at lower transit = 48° 34’ 10” 
ap. zen. dist. of pole [calculated] = 47°. 4. 

Note. The fixed stars are so distant that they suffer no displace- 
ment on account of parallax, appearing in the same direction 
whether seen from the earth’s centre, or from any point on its surface, 

Now, the azimuths and hour angles of a Cygni being observed 
at the successive zenith distances, 15°, 30°, 45°, 60°, 75°, 80°, 
85°, 90°, in the triangle ABC, we have (fig. 135.) 

A= observed hour angle, a, = ob, zen, dis., a = calcul. zen. dis, 

b =cal, zen. dis. of pole = 47° 4’, C, = ob. azimuth, C = cal. az, 

c = polar dis. of star = 45° 15’ 41‘5", [Naut. Al.] 


Barometer 30 inches, internal Thermometer 50°, external 47°. 


(a — ay) : tana, 

00' 1545" 57847 
2 47 5916 |30 30 00 336 00 33‘6 58200 
4 16 1510 45 00 581 00 58*1 58‘100 
5 5l 7°40 60 O1 405 00 405 58‘024 
7 40 53°62 75 03 3453 03 34:3 57421 
8 24 53:48 80 05 20/0 05 200 56'425 
9 17 48:33 65 09 58:0 09 58:0 52318 
0 33 51‘0 33 510 00‘000 


The values of C, are found the same with those of C, which, 
therefore, the student may calculate and verify by employing them 
to find the values of A. 


* The student should find the calculated numbers for himself. 


310 PARALLAX. 


It appears from the above that the effect of the atmospheric re- 
fraction is to elevate a heavenly body, without changing its azi- 
muth, by an angle which varies, to within 10° or 15° of the horizon, 
nearly as the tangent of the zenith distance. With this limitation, 
therefore, we may assume 

(a—a,) : tana, = 58" 
a= da,-+ 58" tana, (640) 

But the refraction is found to vary with the density of the atmo- 
sphere, at 45° zenith distance increasing about 2” for every addi- 
tional inch of the barometer [B] above 30 inches [B— 30], and 
decreasing ‘12" for each additional degree of Fahrenheit’s ther- 
mometer over 50° [ T’— 50]; hence (640) becomes 

a = a,+ [58 + 2(B — 30) —‘12( T — 50) |" tana,. (641) 

Scholium. On account of the fluctuating nature of the atmo- 
sphere, observations within 15° or 20° of the horizon should be 
avoided. For the variation of the magnetic needle, however, the 
azimuth of the sun may be taken when the lower limb appears 
above the horizon by 2 its diameter. 


IV. For the parallaz. Let 8 be the posi- 
tion of a heavenly body seen from a point, 
A, of the earth’s surface, O the centre, Z 
the zenith, and 8, the position of the same 
body in the horizon ;* let 7 ASO =p, ZAS 
=a, ZO08 +a, ; AS,0 = pj, LAS; = 90° % 
then p, the difference of a and a,, is called 
the parallax in altitude, p, the horizontal parallaz, and we have, 
putting OS =d, OA =7, 

sinp : sina=r : d, andsinp, : sin90°=r : d; 


Fig. 140. 


sinp = sinp, sind, or p = p, sina, (642) 
when the parallaxes are small. 
ad, = a—p, sina, (643) 


Scholiwm. The horizontal parallaxes are given in the Nautical 
Almanac. 

V. For the semidiameter. Let S be the point of contact of the 
line SA and a heavenly body, as the sun, whose centre denote by 
S,.; there results, 


nGAR SEA eS. SAS [ = F da 


sina 


+ These positions are supposed to be corrected for refraction. 


SEMIDIAMETER, 311 


SS, sind | 


or tan semidi. s = tan SAS, = . : 
d__ sina, 


but, if we denote by s, what s becomes when the point S falls a 
little below S, so as to make Z SAO = SOA (in which case it is 
obvious that s, may be taken for the horizontal semidiameter), we 
have 


i <! 
c 


tans, = 7? [sina = sind,. | 
sina 
tans = tans, » — ’ 
sind, 


vee sin 
or semidiameter s= hor. sem. s, « 
sin 


a 
ao nearly. (644) 
In (643), replacing a, = ZOS, and a by ZOS, and a@—s, we find 
ZOS, = a —s —p, sin(a — s), = a, — s, nearly. (645) 
Scholium I, The refraction for the ray S,A is less than that for 
SA; therefore, denoting this difference by dif, ref. (S,,S), when 
great accuracy is required, as in finding the longitude, (645) be- 
comes 
True zen, dist, ZOS,=a—s—p, sin(a — s)-+dif. ref. (S,,8). (646) 
Scholium II, Except for the moon, we may assume (647) 
§ = &. . 
Example. On the 20th of August, 1847, at 6h. Greenwich mean 
time, the eye of the observer being 24 feet above the level of the 
sea, the altitude of the moon’s lower limb above the apparent hori- 
zon was found to be 40° 20’. At the same time the barometer in- 
dicated 29 inches, and the thermometer 40°, Required the true 
geocentric zenith distance of the moon’s centre. 


40° 20' 000” 
d (638) 4 49+3" 
Ap. alt. from tr. horizon, 40° 15’ 10697” 
Ap. zen. dis. of moon’s lower limb, 49° 44’ 49+3” 
Refraction (641) I’ 76" 
True zen. dis., a, of moon’s lower limb, 49° 45’ 569” 


Parallax, p, (642), [p, = 56’ 245” N, A.] 43° °3°7" 
Tr. geocentric zen. dis, of moon’s lr. limb a,, 49° 2° 53*2” 
Semidiameter, s, (644) [s, = 15' 22:3" N. A.] 15° 35:0" 
True geocentric zen, dis. of moon’s centre 48° 47’ 16:2" 


The student may make further corrections (646), 


ADDENDUM I. 


The ingenious theorem contained in the following communication 
is published by permission. G. C. W. 


Ricumonp, June 22d, 1848. 
Proressor WHITLOCK, 

Dear Sir:—I have modified the demonstration of my theorem 
as you requested, and hasten to send you the result, 

I shall, at the earliest opportunity, attend to the problem which 
you suggested, to find the area of a polygon in terms of its sides 
and angles, excepting two sides and one angle. If I arrive at 
anything of importance, I will send it to you. 

Yours respectfully, 
Dascum GREEN. 


To find a general expression for the area of a polygon. 


Keté(a, 0; epee, 75 #1) 
be any polygon, Produce 
the sides b, c,..., 7, 7, till 
they meet J, or / produced, 
and denote the correspond- 
ing productions by Jb’, c, 

Sie the expression for 
the double area of a triangle (a, b, c), in terms of its three angles 
and one side is 


a? sin(a,b) sin(a,c) 
sin(d,c) 
we shall have, if we designate the area of the polygon a, 5, ..., d, 
by P, 


POLYGON, 313 


free (j +7)? sin(ky) sin(y,2) _ (¢+- 7)? sin(j,7) sin(?,/) 


sin(d,k) “sin( 7,2) et 
(b+.8')? sin(c,d) sin(d,2) a? sin(,a) sin(a,/) 
4 sin(c,/) i sin(d,/) 
__ a’ sin(a,b) sin(a,/) , (6+ 0’)? sin(d,c) sin(d,/) 
7 sin(,/) Wah ian A). Ayjstatend) 
(c+ c')? sin(c,d) sin(c,2) 
ny ain(d,l) aj 
tel Pok Faint tea nOy 
sin(k,/) 


in which it remains to determine J’, c’, ..., 7’. 


Waiters bo ue sin(a,/) y 2 sin(a,/) | 


Go we Oy ay | TID) A 
c!__sin(b,l) |, _ (B+ B/) sin(d,! 
b'-bohipy seine, laa aro. sin(c,/) 
SEY POU a Cia gin 6 
sin(c,/) 


Po, SEG 
itd sin(j,l)° 
. _( +7) sin(Z, Yikes a sin(a,l) +5 sin(d,l) +.. sed sin(?, ty 
sin(7,) sin(7,/) 
Substituting these values, we obtain 


a sin(a,/) : 
a? sin(a,b) sin(a,/) [% ~sin(6,l) iF sin(d,c) sin(d,2) 
sin(0,/) T sin(c,/) 


jl) +5 sin(d,/ : 
HN Sai PN ected ce ALS ial id kal oe onl) ain art sin(c,d) sin(c,/) 


sin(d,/) Br 


a sin(a,/)-+b sin(d,/)+...4¢ sin(z,/)]?. Ce 
LY Be kaw a eae sin( 7,4) sin( js!) 
sin(k,]) ’ 
sin(a,b) 
sin(a,l) sin(d,/) 


2P= 


*, 2P =[a sin(a,/)]* « 


sin(d,c) 


Shee art aE SUIALET NY 35005, 1) win (G50) 


314 POLYGON. 


sin(c,d) 
* sin(e,) sin(dl) 7" 
sin( 7, k) 
sin( 7,2) sin(A,l)’ 
which is the expression required, and which embraces the sides 
a, b, c, ..., 9, and the angles (a,/), (0,2), (c,l),...5 (Kl); (ad), (b,c), 
(2,0) 7 wit ann) 


+a sin(a,l) + b sin(d,l) + ¢ sin(c,/)}* 


+ [a sin(a,/)+- sin(d,/)-+...+-7 sin(7,/) |? 


ADDENDUM II. 


The student who may desire a partial course in geometry and 
trigonometry, may omit all after page 108 to p. 170, excepting pp. 
145, 153, 154, 155, 156, 157, 158, 159, which are to be read, and 
substitute for Propositions V, VI, VII, VIII, on pages 172, 173, 
174, 175, the following demonstration, after which pages 176, 177, 
178, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200 will be read, 
and then the Book on surveying. 

Let a, b, be two adjacent ares of 
the same circumference, and a+b 
<90°; draw sina intercepting 
cosa, sind intercepting cosd, sin 
(a+6)=m-+n, m being a per- 
pendicular from the foot of sind 
upon cose, and n the portion of 
sin(a-+ b) above a parallel to cosa 
drawn through the foot of sind; 
r being the radius of the circle, 
from similar triangles we have the 
proportions, 


mm _ sina ; sina cosb 
SS 9 s e m = SS ee 9 
cosb Tr T 
nm cosa | cosa sinb 
sind 1% edge o , ? 


hits ieg en ed dyes sina cosb nt cosa sinb , 
or, when r=1, sin(a+ )) =sina cosb-+ cosa sinb ; 
age sin[a + (— b)] = sina ‘cos(— 6) + cosa sin(— 5), 
or (180), (306), sin(a@— 6) =sina cosh — cosa sinb ; 
. (302), cos(a+ b) = sin[ 90° — (a+b) ] 
= sin[ (90° — a) — }] 
= sin(90° — a) cosb — cos(90° — a)sind 
= cosa cosh — sina sind; 
.". (180), (306), | cos(a— b)= cosa cosh + sina sinb, 
and (320) is demonstrated. 


LOGARITHMS OF NUMBERS. 


y= yt0+P'42(0-1)- 


1 e 


[DOU RHA WOH S| 


[> CoOImapwrrHo| 


WAIHI Wee o} 


100 
0000000 
0043214 
0086002 
0128372 
0170333 
0211893 
0253059 
0293838 
0334238 
0374265 

170 


110 
0413927 
0453230 
0492180 
0530784 
0569049 
0606978 
06445890 
0681859 
0718820 
0755470 

180 


2304489 
2329961 
2355284 
2380461 
2405492 
2430380 
2455127 
2479733 
2504200 
2528530 
240 
3802112 
3820170 
3838154 
3856063 
3873898 
3891661 
3909351 
3926970 
3944517 
3961993 


2552725 
2576786 
2600714 
2624511 
2648178 
2671717 
2695129 
2718416 
2741578 
2764618 
250 
3979400 
3996737 
4014005 
4031205 
4048337 
4065402 
4082400 
4099331 
4116197 
4132998 


120 
0791812 
0827854 
0863593 
0899051 
0934217 
0969100 
1003705 
1038037 
1072100 
1105897 


1172713 
1205739 
1238516 
1271048 
1303338 
1335389 
1367206 
139879 1 
1430148 


Pe 4 (2 —1)(2—2) + 


D; 


1 cQaei3” 


130 
1139434 | 1461280 


140 


1492191 
1522883 
1553360 
1583625 
1613680 
1643529 
1673173 
1702617 
1731863 


190 


200 


2787536 
2810334 
2833012 
2855573 
2878017 
2900346 
2922561 
2944662 
2966652 
2988531 
260 
4149733 
4166405 
4183013 
4199557 
4216039 
4232459 
4248816 
4265113 
4281348 
4297523 


3010300 
3031961 
3053514 
3074960 
3096302 
3117539 
3138672 
3159703 
3180633 
3201463 

270 
4313638 
4329693 
4345689 
4361626 
4377506 
4393327 
4409091 
4424798 
4440448 
4456042 


3222193 
3242825 
3263359 
3283796 
3304138 
3324385 
3344538 
3364597 
3384565 
3404441 
280 
4471580 
4487063 
4502491 
4517864 
4533183 
4548449 
4563660 
4578819 
4593925 
4608978 


210. | 


150 
1760913 
1789769 
1818436 
1846914 
1875207 
1903317 
1931246 
1958997 
1986571 
2013971 

220 
3424227 
3443923 
3463530 
3483049 
3502480 
3521825 
3541084 
3560259 
3579348 
3598355 

290 


4623980 


4638930 
4653829 
4668676 
4683473 
4698220 
4712917 
4727564 
4742163 
4756712 


160 


2041200 

2068259 
2095150 
2121876 
2148438 
2174839 
22101081 
2227165 
2253093 
2278867 

230 

3617278 

3636120 
3654880 
3673559 
3692159 
3710679 
3729120 
3747483 
3765770 
3783979 

300 

4771213 

4785665 
4800069 
4814426 
4828736 
4842998 
4857214 
4871384 
4885507 
4899585 


LOGARITHMS 


OF NUMBERS, 


31 


7 


[CDI FAB wWOR S| 


|oDIMABEVH S| [ODIMMABYH S| [CDYRMURwWHHO| 


[CODIMTAWDH Oo 


310 
4913617 
4927604 
4941546 
4955443 
4969296 
4983106 
4996871 
5010593 
5024271 
5037907 
320 
5797836 
5809250 
5820634 
5831988 
5843312 
5854607 
5865873 
5877110 
5888317 
5899496 
450 
6532125 
6541765 
6551384 
6560982 
6570559 
6580114 
6589648 
6599162 
6608655 
6618127 
520 
7160033 
7168377 
7176705 
7185017 
7193313 
7201593 
7209857 
7218106 
7226339 
7234557 
590 
“708520 
1715875 
77123217 
717130547 
1737864 
7745170 
7752463 
77159743 
7767012 
77714268 


320 
5051500 
5065050 
5078559 
5092025 
5105450 
5118834 
5132176 
5145478 
5158738 
5171959 


~~ 390 


400 


340 
5314789 
5327544 
5340261 
535294 | 
5365584 
5378191 
5390761 
5403295 
5415792 
5428254 


330 
5185159 
5198280 
5211381 
5224442 
5237465 
5250448 
5263393 
5276299 
5289167 
5301997 


5910646 
5921768 
5932861 
59435926 
5954962 
5965971 
5976952 
5987905 
5998831 
6009729 


460 


6627578 


6637009 
6646420 
6655810 
6665180 
6674530 
6683859 
6693169 
6702459 
6711728 
530 


6020600 
6031444 
604226 | 
6053050 
6063814 
6074550 
6085260 
6095944 
6106602 | 6211763 
6117233 | 6222140 
470 480 
6720979 | 6812412 
6730209 | 6821451 
6739420 | 6830470 
6748611 | 6839471 
6757783 | 6848454 
6766936 | 6857417 
6776070 | 6866363 
6785184 | 6875290 
6794279 | 6884198 
6803355 | 6893089 


~ 540 550 


6127839 
6138413 | 
6148972 | 
6159501 
6170003 
6180481 
6190933 
6291361 


7242759 
7250945 
7259116 
7267272 
7275413 
7283533 
7291648 
7299743 
7307823 
7315888 


7323938 | 7403627 
7331973 | 7411516 
7339993 | 7419391 
7347998 | 7427251 
7355989 | 7435098 
7363965 | 7442930 
7371926 | 7450748 
7379873 | 7458552 
7387806 | 7466342 
7395723 | 7474118 


ay 410° tor 


350 
5440680 
5453071 
5465427 
5477747 
5490033 
5502284 
5514500 
5526682 
5538830 
5550944 
420 
6232493 
6242821 
6253125 
6263404 
6273659 
6283889 
6294096 
6304279 
6314438 
6324573 


360 
5563025 
5575072 
5587086 
5599966 
5611014 
5622929 
5634811 
5646661 
5658478 
5670264 


370 
5682017 
5693739 
5705429 
5717088 
5728716 
5740313 
5751878 
5763414 
5774918 


5786392 |! 


430 
6334685 
6344773 
6354837 
6364879 
6374897 
6384893 
6394865 
6404814 
6414741 
6424645 


440 
6434527 
6444386 
6454223 
6464037 
6473830 
6483600 
6493349 
6503075 
6512780 


6522463 | 


490 
6901961 
6910815 
6919651 
6928469 
6937269 
6946052 
6954817 
6963564 
6972293 
6981005 

560 
7481880 
7489629 
7497363 
7505084 
7512791 
7520484 
7528164 
7535831 
7543483 
7551123 


600 
7781513 
T7188745 
7795965 
7803173 
7810369 
7817554 
7824726 
7831887 
7839036 

7846173 


610 620 


630. 


500 


6989700 
6998377 
7007037 
7015680 
7024305 
7032914 
7041505 
7050080 
7058637 
7067178 
570 
7558749 
7566061 
7573960 
7981546 
7589119 
7596678 
7604225 
7611758 
7619278 


510 
7075702 
7084209 
7092700 
TLO11L74 
7109631 
7118072 
7126497 
7134905 
7143298 
7151674 

580 
7634280 
7641761 
7649230 
7656686 
7664128 
7671559 
7678976 
768638 L 
7693773 
7701153 


650 


7853298 | 7923917 
7860412 | 7930916 
7867514 | 7937904 
7874605 | 7944880 
7881684 | 7951846 
7888751 | 7958800 
7895807 | 7965743 
7902852 | 7972675 
7909885 | 7979596 
7916906 | 7986506 


7993405 
8000294 
8007171 
8014037 
8020893 
8027737 
8034571 
8041394 
8048207 
8055009 


8061800 
8068580 
8075350 
8082110 
8088859 
8095597 
8102325 
8109043 
8115750 
8122447 


8129134 
8135810 
8142476 
8149132 
8155777 
8162413 
8169038 
8175654 
8182259 
8188854 


318 LOGARITHMS OF NUMBERS, 


660 670 680 690 700 710 70 || 
8195439 | 8260748 | 8325089 | 8388491 | 8450980 | 8512583 | 8573325 
8202015 | 8267225 | 8331471 | 8394780 | 8457180 | 8518696 | 8579353 
8208580 | 8273693 | 8337844 | 8401061 | 8463371 | 8524800 | 8585372 
8215135 | 8280151 | 8344207 | 8407332 | 8469553 | 8530895 | 8591383 
8221681 | 8286599 | 8350561 | 8413595 | 8475727 | 8536982 | 8597386 
8228216 | 8293038 | 8356906 | 8419848 | 8481891 | 8543060 | 8603380 
8234742 | 8299467 | 8363241 | 8426092 | 8488047 | 8549130 | 8609366 
8241258 | 8305887 | 8369567 | 8432328 | 8494194 | 8555192 | 8615344 
8247765 | 8312297 | 8375884 | 8438554 | 8500333 | 8561244 | 8621314 
8254261 | 8318698 | 8382192 | 8444772 | 8506462 | 8567289 | 8627275 
730 740 750 760 77 780 790 
8633229 | 8692317 | 8750613 | 8808136 | 8864907 | 8920946 | 6976271 
8639174 | 8698182 | 8756399 | 8813847 | 8870544 | 8926510 | 8981765 
8645111 | 8704039 | 8762178 | 8819550 | 8876173 | 8932068 | 8987252 || 
8651040 | 8709888 | 8767950 | 8825245 | 8881795 | 8937618 | 8992732 |. 
B656961 | 8715729 | 8773713 | 8530934 | 8987410 | 8943161 | 8998205 
8662873 | 8721563 | 8779470 | 8836614 | 8893017 | 8948697 | 9003671 || 
8668778 | 8727388 | 8785218 | 8842288 | 8898617 | 8954225 | 9009131 
8674675 | 8733206 | 8790959 | 8817954 | 8904210 | 8959747 | 9014583 
8680564 | 8739016 | 8796692 | 8853612 | 8909796 | 8965262 | 9020029 
8686444 | 8744818 | 8802418 | 8859263 | 8915375 | 8970770 | 9025468 
800 810 820 830 840 850 860 
9030900 | 9084850 | 9138139 | 9190781 | 9242793 | 9294189 | 9344985 
9036325 | 9090209 | 9143432 | 9196010 | 9247960 | 9299296 | 9350032 
9041744 | 9095560 | 9148718 | 9201233 | 9253121 | 9304396 | 9355073 
9047155 | 9100905 | 9153998 | 9206450 | 9258276 | 9309490 | 9360108 
9052561 | 9106244 | 9159272 | 9211661 | 9263424 | 9314579 | 9365137 
9057959 | 9111576 | 9164539 | 9216865 | 9268567 | 9319661 | 9370161 
9063350 | 9116902 | 9169800 | 9222063 | 9273704 | 9324738 | 9375179 
9068735 | 9122221 | 9175055 | 9227255 | 9278834 | 9329808 | 9380191 
9074114 | 9127533 | 9180303 | 9232440 | 9283959 | 9334873 | 9385197 |: 
9079485 | 9132839 | 9185545 | 9237620 | 9289077 | 9339932 | 9390198 || 
870 880 890 900 910 920 930 
9395193 | 9444827 | 9493900 | 9512425 | 9590414 | 9637878 | 9684829 
9400182 | 9449759 | 9498777 | 9547248 | 9595184 | 9642596 | 9689497 
9405165 | 9454686 | 9503649 | 9552065 | 9599948 | 9647309 | 9694159 
9410142 | 9459607 | 9508515 | 9556878 | 9304708 | 9652017 | 9698816 
9415114 | 9464523 | 9513375 | 9561684 | 9609462 | 9656720 | 9703469 
9420081 | 9469433 | 9518230 | 9566486 | 9614211 | 9661417 | 9708116 
9425041 | 9474337 | 9523080 | 9571282 | 9618955 | 9666110 | 9712758 
9429996 | 9479236 | 9527924 | 9576073 | 9623693 | 9670797 | 9717396 
9434945 | 9484130 | 9532763 | 9580858 | 9628427 | 9675480 | 9722028 
9439889 | 9489018 | 9537597 | 9585639 | 9633155 | 9680157 | 9726656 
940 950 960 970 980 990 1000 
9731279 | 9777236 | 9822712 | 9867717 | 9912261 | 9956352 | 0000000 
9735896 | 9781805 | 9827234 | 9872192 | 9916690 | 9960737 
9740509 | 9786369 | 9831751 | 9876663 | 9921115 | 9965117 
9745117 | 9790929 | 9836263 | 9881128 | 9925535 | 9969492 
9749720 | 9795484 | 9840770 | 9885590 | 9929951 | 9973864 
9754318 | 9200034 | 9845273 | 9390046 | 9934362 | 9978231 
9758911 | 9804579 | 9849771 | 9894498 | 9938769 | 9982593 
9763500 | 9809119 | 9854265 | 9898946 | 9943172 | 9986952 
9768083 | 9813655 | 9858754 | 9903389 | 9947569 | 9991305 
9772662 | 9818186 | 9863238 | 9907827 | 9951963 | 9995655 


| CHIMUR WWHO| 


] Pau mMIRwWWR S| 


| CDRA TR WOH S| 


| OLIATRWWR S| 


—— 


[oma ms aRwwrs| 


LOGARITHMIC SINES. 


D, D, D, 
Yo=Yotss LER Aare Nr 5 1 (rt — 1)(t — 2) vias art 


From 0*1° to 0‘5° inclusive the characteristic is 3, from 0‘6° to 


5‘7° it is 2, and after that I. 


| ge joe if go a 3° 49° 5e° 6° Hh 


‘(0} —infin.| 2418553} 5428192; 7188002) 8435845] 9402960| 0192346 
| ‘1'| 2418771 | 2832434] 5639994] 7330272! 8542905] 9488739! 0263865 | 
21} 5429065} 3210269| 5841933} 7468015} 8647376] 9572843! 0334212 
$8} 7189966! 3557835!] 6034886! 7601512] 8749381] 9655337] 0403424 
‘4/ 8439338] 3879622| 6219616/| 7731014} 8849031 | 9736280 | 0471538 
‘| 9408419| 4179190} 6396796| 7856753} 8946433 | 9815729] 0538588 
‘6 | 0200207} 4459409] 6567017! ‘79738941 | 9041685! 9893737] 0604604 
‘7| 0869646! 4722626} 6730804; 8097772} 9134881 | 9970356; 0669619 
18] 1449532] 4970784| 6888625; 8213425] 9226105| 0045634 733663 |}. 
‘9 | 1961020) 5205514} 7040899} 8326066) 9315439} 0119616| 0796762 
| 7° xo go 10° 11° it ee eB 
(6) | 08589145} 1435553} 1943324) 2396702; 2805988} 3178789} 3520880 
1] 0920237] 1489148] 1990913] 2439472] 2844803] 3214297] 3553582 
2] 0980662 | 1542076 | 2037974) 2481811 | 2883260] 3249505! 3586027 
3} 1040246 | 1594354] 2084516 | 2523729) 2921367] 3284416 | 3618217 
| 
| 


141 1099010] 1645998} 2130552] 2565233 | 2959129} 3319035) 3650158 
(5) | 1156977) 1697021 | 2176092) 2606330} 2996553} 3353368] 3681853 |; 
6 | 1214167) 1747439 | 2221147} 2647030] 3033644] 3387418) 3713304 
ie7 | 1270600] 1797265} 2265725 | 2687338 | 3070407) 3421190} 3744517 
8} 1326297} 1846512} 2309838] 2727263} 3106849} 3454688) 3775493 
9 | 1381275] 1895195 | 2353494 | 2766811 | 3142975} 3487917] 3806237 


. 14° io? ed oy. is Re ie 20° 
0 | 3836752) 4129962} 4403381) 4659353 | 4599824] 5126419] 5340517 
‘1 | 3867040] 4158152] 4429728} 4681069 | 4923083] 5148371] 5361286 
12! 3897106} 4186148] 4455994; 4708631 | 4946205) 5170198} 5381943 
3 | 3926952 | 4213950} 4481909) 4733043 | 4969192] 5191904] 5402489 
‘4| 3956581 | 4241563} 4507747) 4757304 | 4992045| 5213488] 5422926 
‘| 3985996 | 4268988} 4533418) 4781418) 5014764| 5234953] 5443253 
«6 | 4015201 | 4296228} 4558926] 4805385 | 5037353} 5256298] 5463472 
7 | 4044196 | 4323285] 4584271) 4829208} 5059818 | 5277526] 5483585 
'|-8| 4072987| 4350161 | 4609456) 4852888} 5082141] 5298638) 5503592 
9 | 4101575 | 4376859 | 4634483| 4876426 | 5104343 | 5319635] 5523494 


a, nn 
aaa 


i te 


22° 


5543292 
5562987 
5582579 
5602071 
5621462 
5640754 
56059948 
5679014 
5698043 
5716946 


5735754 
5754468 
5773088 
5TIL616 
5810052 
5828397 
5346651 
5864816 
5482892 
5900830 


LOGARITHMIC SINES. 


23° 
5918780 
5936594 
5954322 
5971965 
5989523 
6006997 
6024388 
6041696 
6058923 
6076063 


28° 


29° 


30° 


24° 


6093133 
6110118 
6127023 
6143850 
6160599 
6177270 
61933864 
6210382 
6226824 
6243190 
Hie. 


6716093 
6730319 
6744485 
6758592 
6772640 
6786629 
6800560 
6814434 
6328250 
6842010 
35° 


6855712 
6869359 
6832949 
6896484 
6909964 
6923388 
6930758 
6959074 
6963336 
6976545 


36° 


6989700 
7002802 
7015852 
7028249 
7041795 
7054689 
7067531 
7080323 
7093063 
7105753 
37° 


75859 13 
7596718 
7607483 
7618208 
7628894 
7639540 
7650147 
7660715 
7671244 


7681735 


42° 
8255109 
8263512 
8271887 
8280231 
8288547 
8296833 
830509 L 
8313320 
8321519 
832969 | 


7092187 
7792601 
7712976 
7723314 
7733614 
7743876 
7754101 
7164289 
7774439 
7784553 
43° 
8337833 
8345948 
8354033 
8362091 
8370121 
8378122 
8386096 
8394041 
8401959 
8409850 


49° 
8777799 
8784376 
8790930 
8797462 
8803970 
BR10455 
8816918 
8823357 
8829774 
8836168 


50° 
8842540 
8845889 
8855215 
8861519 
8867801 
8374061 
8880298 
8886513 
8892706 
8898877 


7794630 
7804671 
7814675 
7824643 
7834575 
7344471 
7854332 
7864157 
7873946 
788370 1 


44° 


8417713 
8425548 
8433356 
8441137 
844889 | 
8456618 
8464318 
8471991 
8479637 
8487257 


7118393 
7130983 
7143524 
7156015 
7168458 
7180851 
7193196 
7205493 
7217742 
7229943 
38° 
7893420 
7903104 
7912754 
7922369 
7931949 
7941496 
7951008 
7960436 
7969930 
7979341 


bi itip ra 


8494850 
8502417 
8509957 
8517471 
8524959 
8532421 
8539856 
8547266 
8554650 
8562008 


25° 
6259483 
6275701 
6291845 
6307917 
6323916 
6339844 
6355699 
6371484 
6387199 
6402844 


26° 
6418420 
6433926 
6449365 
6464735 
6480038 
6495274 
6510444 
6525548 
6540586 
6555559 


a7° || 
6570468 
6585312 
6600093 
6614810, 
6629464 
6644056 | 
6658586 
6673054 | 
6687461 
6701807 


g 32° 


33° 


34° 


7242097 
7254204 
7266264 
7278277 
7290244 
7302165 
7314040 
7325870 
7337054 
7349393 
39° 
7988718 
7998062 
8007372 
5016649 
5025894 
8035105 
8044284 
8053430 
8062544 
8071626 


45°h 


7361088 
7372737 
7384343 
7395904 
7407421 
7418895 
7430325 
7441712 
7453056 
7464358 
49° 
8080675 
8089692 
8098678 
8107631 
8116554 
8125444 
8134303 
8143131 
8151928 
8160694 


7475017 | 
7436833 

7498007 
7509140 
71520231 
7531280 |! 
7542288 
7553256 
7564182 
7575068 

41° 

8169429 
8178133 
8186807 
8195450 
8204063 | 
8212646 
8221198 
8229721 || 
8238213 

8246676 


47° 


UE ARO i 


8569341 
8576648 
8583929 
8591186 
8598416 
8605622 
8612803 
8619958 
8627088 
8634194 


8641275 
8648331 
8655362 
8662369 
8669351 
8676309 
8683242 
8690152 
8697037 
8703898 


Dre 


8905026 
8911153 
8917258 
8923342 
8929404 
8935444 
8941462 
8947459 
8953435 
8959389 


52° 
8965321 
8971233 
8977123 
8982992 
8988840 
8994667 
9000472 
9006257 
9012021 
9017764 


9074059 | 9128328 | 9180620 


53° 
9023486 
9029188 
9034868 
9040529 
9046168 
9051787 
9057386 
9062964 
9068522 


54° 
9079576 
90850738 
9090550 
9096007 
9101444 
9106860 
9112257 
9117634 
9122991 


ge 


8710735 
8717548 
8724337 
8731102 
8737844 
8744561 

8751256 

8757927 
8764574 
8771198, 


9133645 
9138943 
9144221 
9149479 | 
9154718 
9159937 
9165137 | 
9170317 
9175478 


LOGARITHMIC SINES. 


56° 


~ 9185742. 


9190845 
9195929 
9200994 
9206039 
9211066 
9216073 
9221062 
9226032 
9230982 


63° 
9498809 


9502663 
9506500 
9510320 
9514124 
9517912 
9521683 
9525437 
9529175 
9532897 


7° 
9729858 


9732610 
9735346 
9738067 
9740774 
9743466 
9746142 
9748804 
9751451 


9754083 
one 
9887239 


9888982 
9890711 
9892427 
9894128 
9895815 
9897489 
9899148 
9900794 


9902426 
84° 
9976143. 


9976933 
9977710 
9978473 
9979223 
9979960 
9980683 
9981393 
9982089 
9982772 


57° 


9235914 


9240827 
9245721 
9250597 
9255454 
9260292 
9265112 
9269913 
9274695 


9279459 


64° 


9536602 


9540291 
9543963 
9547619 
9551259 
9554882 
9558490 
9562081 
9565656 
9569215 


ai? 
9756701 


9759303 
9761891 
9764464 
9767022 
9769566 
9772095 
9774609 
9777108 
9779593 


78° 
9904044 


9905648 
9907239 
9908815 
9910378 
9911927 
9913462 
9914984 
9916492 
‘9917986 


oe 
9983442 


9984099 
9984742 
9985372 
9985988 
9986591 
9987181 
9987758 
9988321 
9988871 


9284205 


65° 
9572757 


72° 


79° 
9919466 


ae 
9989408 


58° 


9288932 
9293641 
9298332 
9303004 
9307658 
9312294 
9316911 
9321511 
9326092 


9576284 
9579794 
9583288 
9586767 
9590229 
9593675 
9597106 
9600520 
9603919 


~ 9782063 
9784519 
9786960 
9789386 
9791798 
9794195 
9796578 
9798946 
9801299 
9803639 


9920932 
9922385 
9923824 
9925250 
9926661 
9928059 
9929444 
9930814 
9932171 


9989931 
9990441 
9990938 
9991422 
9991892 
9992349 
9992793 
9993223 
9993640 


59° 
~ 9330656 
9335201 
9339729 
9344238 
9348730 
9353204 
9357660 
9362098 
9366519 
9370921 


66° 
9607302 


9610668 
9614020 
9617355 
9620674 
9623978 
9627266 
9630538 
9633795 
9637036 


73° 
9805963. 


9808273 
9810569 
9812850 
9815117 
9817370 
9819608 
9821831 
9824041 
9826236 


B0° 
9933515 


9934844 
9936160 
9937463 
9938752 
9940027 
9941289 
9942537 
9943771 
9944992 


egieT? Pie 
9994044 


9994435 
9994812 
9995176 
9995527 
9995865 
9996189 
9996500 
9996798 
9997082 


21 


TY 60° 


~ 9375306 


9379674 
9384024 
9388356 
9392671 
9396968 
9401248 
9405510 
9409755 
9413982 


67° 
9640261 


9643470 
9646665 
9649843 
9653006 
9656153 
9659285 
9662402 
9665503 
9668588 


9830583 
9832735 
9834872 
9836996 
9839105 
9841200 
9843281 
9845347 
9847400 


9946199 
9947393 
9948573 
9949740 
9950893 
9952033 
9953159 
9954271 
9955370 
9956456 


sine BOON oe: 
9997354 


9997612 
9997856 
9998088 
9998306 
9998512 
9998703 
9998882 
9999047 
9999200 


maT ele 
9828416 


81° 


61° 
9418193 
9422386 
9426561 
9430720 
9434861 
9438985 
9443092 
9447182 
9451255 
9455310 


68° 
9671659. 


9674713 
9677753 
9680777 
9683786 
9686779 
9689757 
9692720 
9695668 
9698600 


i a 
9849438 


9851462 
9853471 
9855467 
9857449 
9859416 
9861369 
9863308 
9865233 
9867144 


82° 
9957528 


9958586 
9959631 
9960663 
9961681 
9962686 
9963077 
9964655 
9965619 
9966570 


8go 
9999338 


9999464 
9999577 
9999676 
9999762 
9999835 
9999894 
9999940 
9999974 
9999993 


321 


| 62° ca 


<a 
9463371 
9467376 
9471364 
9475335 
9479289 
9483227 
9487147 
9491051 
9494938 


A as 


~ 9701517 
9704419 
9707306 
9710178 
9713035 
9715876 
9718703 
9721514 
9724310 
9727092 


76° 
9869041 


9870924 
9872793 
9874648 
9876488 
9878315 
9880128 
9881927 
9883712 
9885482 


83° 


~ 9967507 
9968431 
9969342 
9970239 
9971122 
9971993 
9972850 
9973693 
9974523 
9975340 


ha ae 
10000000 


LOGARITHMIC TANGENTS, 


ae D, D, 7. vt 
Sere eM apeh ehiell) tires take Dae) Mesa ae 


From 01° 0‘6° 58° 45° 843° 89‘5° 
To 05° BI7° , 449° 84:29 894° 89°9° 
Char. 3 2 1 0 1 2 


6° 
0216202 
0288524 
0359688 
0429731 
0498689 
0566595 
‘0633482 
0699381 
0764321 
0828331 

13° 
3633641 
3668100 
3702315 
3736291 
3770030 
3803537 
3836816 
3869869 
3902700 
3935313 
20° 

5610659 
5634194 
5657633 
5680975 
5704223 
5727377 
5750438 


5773407 | 
5796286 | 
5819074 


5° 
9419518 
9505967 
9590754 
9673944 
9755597 
9835769 
9914514 
9991383 
0067924 
0142682 

12° 
3274745 
3311872 
3348711 
3385267 
3421546 
3457552 
3493290 
3528763 
3563977 
3598935 
19° 

5369719 
5394287 
5418747 
5443100 
5467346 
5491487 
5515524 
5539459 

5563292 

5587025 


4° 
8446437 
8554034 
8659055 
8761623 
8861850 
8959842 
9055697 
9149509 
9241363 
9331340 
11° 
2886523 
2926817 
2966769 
3006383 
3045667 
3084626 
3123266 
3161592 
3199611 
3237327 
18° 
5117760 
5143490 
5169097 
5194583 
5219950 
5245199 
5270331 
5295347 
5320250 
5345040 


3° 
7193958 
7336631 
7474792 
7608719 
77138665 
71864861 
7987519 
8106834 
8222984 
8336134 
10° 
2463188 
2507301 
2550997 
2594285 
2637173 
2679669 
2721780 
2763514 
2804878 
_ 2345878 
17° 


4853390 
4880430 
4907332 
4934097 
4960727 
4987223 
5013588 
5039822 
5065928 
5091907 


go 
5430838 
5642912 
5845136 
6038386 
6223427 
6400931 
6571490 
6735628 
6893813 
7046465 
ee os 
1997125 
2045922 
2094203 
2141980 
2189264 
2236065 
2282395 
2328262 
2373678 
2418650 
16° 
4574964 
4603492 
4631863 
4660078 
4688139 
4716048 
4743808 
4771421 


1° 
2419215 
2833234 
3211221 
3558953 
3880918 
4180679 
4461103 
4724538 
4972928 
5207902 
noes 
1478025 
1582692 
1586706 
1640083 
1692839 
1744988 
1796546 
1847525 
1897939 
1947802 
15° 
4280525 
4310753 
43 10800 
4370670 
4400363 
4429883 
4459232 
4488413 
4219515 | 4517427] 4798887 
4250113 | 4546276 | 4826210 


| acceraeseaS SGOT SE ie a WS eerie erm 


0000000 
2418778 
5429091 
7190026 
8439444 
9408584 
0200445 
08699'70 
1449956 
1961556 
ic one 
“Q| 0891438 
0953667 
1015044- 
1075591 
1135333 
1194291 


1422689 
14° 
3967711 
3999896 
4031873 
4063644 
4095212 
4126581 
4157752 
4188729 


21° 


0} 5841774 


5864386 
5886912 
5909351 
5931705 
5953975 
5976162 
5998267 
6020290 
6042233 


28° 


7256744 
7275008 
7293230 
7311410 
7329547 
7347644 
7365699 
7383714 
7401689 
7419624 


3n° 


0] 8452268 


8468390 
8484492 
8500575 
8516637 
8532680 
8548704 
8564708 
8580694 
8596661 


42° 


9544374 
9559615 
9574850 
9590080 
9605305 
9620525 
9635740 
9650951 
9666157 
9681360 


49° 


0608369 
0623682 
063902 
0654330 
0669666 
0685011 
0700364 
0715726 
0731096 


0746476 


LOGARITHMIC TANGENTS, 


22° 


6064096 
6085880 
6107586 
6129214 
6150766 
6172243 
6193645 
6214973 
6236227 
6257409 
ae 
7437520 
7455376 
7473194 
7490974 
7508716 
7526420 
7544088 
7561718 
7579313 
7596871 
36° 
8612610 
8628541 
8644454 
8660350 
8676228 
8692089 
8707933 
8723760 
8739571 
8755365 


43° 


9696559 
9711754 
9726945 
9742133 
9757318 
9772500 
9787679 
9802856 
9818030 
9833202 
50° 
0761865 
0777263 
0792671 
0808089 
0823517 
0838955 
0854404 
0869863 
0885334 
0900815 


6278519 


30° 


44° 
~ 9848372 


ot es 


23° 


6299558 
6320527 
6341426 
6362257 
6383019 
6403714 
6424342 
6444903 
6465400 


7614394 
7631881 
71649334 
7666751 
7684135 
7701485 
7718801 
7736084 
7753334 
77710552 
37° 


24° 
6485831 
6506199 
6526503 
6546744 
6566923 
6587041 
6607097 
6627093 
6647030 
6666907 
31° 
7787737 
780489 1 
7822013 
7839104 
7856 164 
7873193 
7890192 
7907161 
7024101 
7941011 


38° 


8771144 
8786907 
8802654 
8818386 
8834103 
8849805 
8865492 
8881165 
8896823 
8912468 


9863540 
9878706 
9893871 
9909035 
9924197 
9939359 
9954520 
9969680 
9984840 


8928098 
8943715 
8959319 
8974910 
8990487 
9006052 
9021604 
9037144 
9052672 
9068188 


45° uA 
0000000 


0015160 
0030320 
0045480 
0060641 
0075803 
0090965 
0106129 
0121294 
0136460 


52° 


1071902 
1087532 
1103177 
1118835 
1134508 
1150195 
1165897 
1181614 
1197346 
1213093 


6686725 


32° 


es 


39° 


46° 


25° 


6706486 
6726190 
6745836 
6765426 
6784961 
6804440 
6823865 
6843236 
6862553 


7957892 
7974745 
7991569 
8008365 
8025133 
8041873 
8058587 
8075273 
8091933 
8108566 


9083692 
9099185 
9114666 
9130137 
9145596 
9161045 
9176483 
9191911 
9207329 
9222737 


0151628 
0166798 
0181970 
0197144 
0212321 
0227500 
0242682 
0257867 
0273055 
0288246 


26° 


6881818 


6901030 
6920189 
6939298 
6958355 
6977363 
6996320 
7015227 
7034086 
7052897 


eet 


8125174 
8141755 
8158311 
8174842 


8191348 | 


8207829 
8224286 
8240719 
8257127 
8273513 


40° 


323 


27° 
7071659 
7090374 
7109041 
7127662 
7146237 
7164767 | 
7183251 
7201690 
7220085 
7238436 
34° 
8289874 
8306213 
8322529 
8338823 
8355094 | 
8371343 
8387571 
8403776 
8419961. 
8436125 


Ser one 


9238135 
9253524 
9268904 
9284274 
9299636 
9314989 
9330334 
9345670 
9360998 
9376318 


47° 


0303441 


0318540 
0333843 
0349049 
0364260 
0379475 
0394695 
0409920 
0425150 


. 0440385 


53° 
1228856 
1244635 
1260429 
1276240 
1292067 
1307911 
1323772 
1339650 
1355546 
1371459 


54° 


1387390 
1403339 
1419306 
1435292 
1451296 
1467320 
1483363 
1499425 
1515508 
1531610 


9391631 
9406936 
9422033 | 
9437524 | 
9452807 
9468084 
9483355 
9498619 | 
9513876. 
9529128 
48° ‘ 
0455626 
0470872 | 
0486124 | 
0501381 | 
0516645 | 
0531916 
0547193 
0562476 | 
0577767 | 
0593064 | 
55° 
1547732 
1563875 
1580039 
1596224 
1612429 
1628657 
1644906 
1661177 | 
1677471 | 
1693787 


324 LOGARITHMIC TANGENTS, 


56° 57° 
| 1710126! 1874826 
1] 1726487} 1891434 
‘Q} 1742873] 1908067 
‘3| 1759281 | 1924727 
‘4) 1775714] 1941413 
5] 1792171! 1958127 
'6| 1808652] 1974867 
| 1825158) 1991635 
3} 1841689} 2008431 
9! 1858245 | 2025255 
63° 64° 
| 2928341 | 3118182 
1] 2947103 | 3137447 
2} 2965914 | 3156764 
‘3| 2984773 | 3176135 
‘4| 3003680 | 3195560 
‘5| 3022637! 3215039 
‘6| 3041645 | 3234574 
7] 3060702 | 3254164 
'«g| 3079811 | 3273810 
‘9| 3098970 | 3293514 
| 70° 71° 
<Q] 4389341 | 4630281 
| 4412975 | 4654960 
‘| 4436708 | 4679750 
‘3] 4460541 | 4704653 
‘4| 4484476 | 4729669 
15 4508513 | 4754801 
«61 45326541 4780050 


62° 
2743256 
2761564 
Q779915- 
2798310 
2816749 
2835233 
2853763 
2872338 | 
2890959 
2909626 
69° 
4158226 | 
4180926 | 
4203714 
4226593 
4249562 
4272623 | 
4295777 | 


61° 
2562480 
2580376 
259831 i 
2616286 
2634301 
2652356 
2670453 
2688590 
2706770 
2724992 
68° 
3935904 
3957767 
3979710 
4001733 
4023838 
4046025 
4068295 
4090649 | 4319025 
4113088 | 4342367 
4135614 | 4365806 


715° 76° 
5719475 | 6032289 
6064687 
6097300 
6130131 
6163184 
6196463 
6229970 
6263709 | 
6297685 | 
6331900 

83° 
9108562 
9171669 
9235679 | 
9300619 
9366518 
9433405 
9501311 
9570269 
8984956 | 9640312 
9046333 | 9711476 

age 90° 
7580785 | + infin. 
8038444 
8550044 
9130030 
9799555 
0591416 
1560556 
2809974 
4570909 
1581222 


58° 59° 60° 
2042108 | 2212263-) 2385606 
2058989 | 2229448} 2403129 
2075899 | 2246666 | 2420687 
2092839 | 2263916 | 2438282 
2109808 | 2281199} 2455912 
2126807 | 2298515} 2473580 
2143836 | 2315865} 2491284 
2160896 | 2333249 | 2509026 
2177987 | 2350666 | 2526806 
2195109 | 2368119 | 2544624 
65° 66° 67° 
3313275 | 3514169 | 3721481 
3333093 | 3534600 | 3742591 
3352970 | 3555097 | 3763773 
3372907 | 3575658 | 3785027 
3392903 | 3596286 | 3806355 
3412959 | 3616981 | 3827757 
3433077 | 3637743 | 3849234 
3453256 | 3658574 | 3870786 
3473497 | 3679473 | 3892414 
3493801 | 3700442] 3914120 
72° 73° 74° 
4882240| 5146610] 5425036 
4908093 | 5173790 | 5453724 
4934072 | 5201113| 5482573 
4960178 | 5228579 | 5511587 | 
4986412 | 5256192| 5540768 
5012777 | 5283952} 5570117 
5039273 | 5311861 | 5599637 
5065903 | 5339922 | 5629330 | 
5092668 | 5368137} 5659200 
5119570| 5396508 | 5689247 
a° 80° 81° 
7113477 | 7536812 | 8002875 
7154122 | 7581350} 8052198 
7195122 | 7626322) 8102061 
7236486 | 7671738 | 8152475 
7278220 | 7717605 | 8203454 
7320331 | 7763935 | 8255012 
7362827 | 7810736 | 8307161 
7405715 | 7858020 | 8359917 
7449003 | 7905797) 8413294 
7492699 | 7954078 | 8467308 
86° 87° 88° 
1553563 | 2806042 | 4569162 
1663866 | 2953535) 4792098 
1777016 | 3106187] 5027072 
1893166 | 3264372] 5275462 
2012481 | 3428510 | 5538897 
2135139 | 3599069 | 5819321 
2261335 | 3776573 | 6119082 
2391281 | 3961614} 6441047 
2525208 | 4154864 | 6788779 
2663369 | 4357088} 7166766 


7} 4556900 | 4805417 
‘§] 4581253 | 4830903 
9} 4605713 | 4856510 
77° 78° 

(| 6366359 | 6725255 
| 6401065 | 6762673 
2} 6436023 | 6800389 
3} 6471237 | 6838408 
‘4| 6506710 | 6876734 
‘5| 6542448) 6915374 
6| 6578454 | 6954333 
‘7| 6614733 | 6993617 
‘8| 6651289 | 7033231 
(9| 6688128 | 7073183 

34° 85° 

‘0! 9783798 | 0580482 
1] 9857318 | 0668660 
2] 9932076 | 0758637 
| 3] 0008117 | 0850491 

‘4} 0085486 | 0944303 
5} 0164231 | 1040158 
6} 0244403] 1138150 
‘7| 0326056 | 1238377 
8} 0409246 | 1340945 
‘9] 0494033 |} 1445966 


6000104 
«82° 
8521975 
8577311 
| 8633335 
8690063 
8747514 
8805709 
8864667 
8924409 


SS 2 
EE CE a ST a a eae TE SST SEMA? 
. *: 


r) 


W x a 
= ' ny 


i) ‘ AG ' -, | 


rv 
Mics 
al 


Nie 
Bert on & i 


‘ag 


via 
habs 
ew) 4 
v U 


+) om ' A hy 
7 ol ww n A 
te a Soro: ; 


4 vi on ‘| . ( 


} 


1) eS ae 
Lae Hed rat 


Meret ure 
ip 


GPetialt Ia) Ne 

SR Ge Lane TAM ND By 
a pat 

oats . Be ts 
Pine 
ft! 


: ; vd ay i : an at 
i ae ri sik a x i 
‘3 " oy | yh ye ne 
te h, ye (e . y" 


' Sia 
igh 


A 


4 iv 


Rol . 
ba 7) 


i 


€ 
iz 


am eS 


+ Mk: 


wig rt 


fab ol) Betoe 


UNIVERSITY OF ILLINOIS-URBANA 


516.2W59E C001 
ELEMENTS OF GEOMETRY, THEORETICAL AND PR 


3.0112 017248235 


20 


